1127 lines
33 KiB
JavaScript
1127 lines
33 KiB
JavaScript
/*
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* rm - Feb 2011
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* ctio-faster.js:
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* This file is a modification of ctio.js that removes all the runtime sanity
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* checking for efficiency.
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*
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* A simple way to read and write simple ctypes. Of course, as you'll find the
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* code isn't as simple as it might appear. The following types are currently
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* supported in big and little endian formats:
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*
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* uint8_t int8_t
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* uint16_t int16_t
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* uint32_t int32_t
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* float (single precision IEEE 754)
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* double (double precision IEEE 754)
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*
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* This is designed to work in Node and v8. It may in fact work in other
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* Javascript interpreters (that'd be pretty neat), but it hasn't been tested.
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* If you find that it does in fact work, that's pretty cool. Try and pass word
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* back to the original author.
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*
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* Note to the reader: If you're tabstop isn't set to 8, parts of this may look
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* weird.
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*/
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/*
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* Numbers in Javascript have a secret: all numbers must be represented with an
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* IEEE-754 double. The double has a mantissa with a length of 52 bits with an
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* implicit one. Thus the range of integers that can be represented is limited
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* to the size of the mantissa, this makes reading and writing 64-bit integers
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* difficult, but far from impossible.
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*
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* Another side effect of this representation is what happens when you use the
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* bitwise operators, i.e. shift left, shift right, and, or, etc. In Javascript,
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* each operand and the result is cast to a signed 32-bit number. However, in
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* the case of >>> the values are cast to an unsigned number.
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*/
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/*
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* A reminder on endian related issues:
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*
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* Big Endian: MSB -> First byte
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* Little Endian: MSB->Last byte
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*/
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/*
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* An 8 bit unsigned integer involves doing no significant work.
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*/
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function ruint8(buffer, endian, offset)
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{
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return (buffer[offset]);
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}
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/*
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* For 16 bit unsigned numbers we can do all the casting that we want to do.
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*/
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function ruint16(buffer, endian, offset)
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{
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var val = 0;
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if (endian == 'big') {
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val = buffer[offset] << 8;
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val |= buffer[offset+1];
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} else {
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val = buffer[offset];
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val |= buffer[offset+1] << 8;
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}
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return (val);
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}
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/*
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* Because most bitshifting is done using signed numbers, if we would go into
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* the realm where we use that 32nd bit, we'll end up going into the negative
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* range. i.e.:
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* > 200 << 24
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* -939524096
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*
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* Not the value you'd expect. To work around this, we instead do a
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* multiplication by (1 << 24) which does the same thing, but in a way that
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* ensures we don't lose that bit.
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*/
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/*
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* Handle the case of losing our MSBit
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*/
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function fixu32(upper, lower)
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{
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return ((upper * (1 << 24)) + lower);
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}
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function ruint32(buffer, endian, offset)
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{
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var val = 0;
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if (endian == 'big') {
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val = buffer[offset+1] << 16;
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val |= buffer[offset+2] << 8;
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val |= buffer[offset+3];
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val = fixu32(buffer[offset], val);
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} else {
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val = buffer[offset+2] << 16;
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val |= buffer[offset+1] << 8;
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val |= buffer[offset];
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val = fixu32(buffer[offset+3], val);
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}
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return (val);
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}
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/*
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* Reads a 64-bit unsigned number. The astue observer will note that this
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* doesn't quite work. Javascript has chosen to only have numbers that can be
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* represented by a double. A double only has 52 bits of mantissa with an
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* implicit 1, thus we have up to 53 bits to represent an integer. However, 2^53
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* doesn't quite give us what we want. Isn't 53 bits enough for anyone? What
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* could you have possibly wanted to represent that was larger than that? Oh,
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* maybe a size? You mean we bypassed the 4 GB limit on file sizes, when did
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* that happen?
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*
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* To get around this egregious language issue, we're going to instead construct
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* an array of two 32 bit unsigned integers. Where arr[0] << 32 + arr[1] would
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* give the actual number. However, note that the above code probably won't
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* produce the desired results because of the way Javascript numbers are
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* doubles.
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*/
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function ruint64(buffer, endian, offset)
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{
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var val = new Array(2);
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if (endian == 'big') {
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val[0] = ruint32(buffer, endian, offset);
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val[1] = ruint32(buffer, endian, offset+4);
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} else {
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val[0] = ruint32(buffer, endian, offset+4);
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val[1] = ruint32(buffer, endian, offset);
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}
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return (val);
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}
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/*
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* Signed integer types, yay team! A reminder on how two's complement actually
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* works. The first bit is the signed bit, i.e. tells us whether or not the
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* number should be positive or negative. If the two's complement value is
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* positive, then we're done, as it's equivalent to the unsigned representation.
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*
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* Now if the number is positive, you're pretty much done, you can just leverage
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* the unsigned translations and return those. Unfortunately, negative numbers
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* aren't quite that straightforward.
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*
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* At first glance, one might be inclined to use the traditional formula to
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* translate binary numbers between the positive and negative values in two's
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* complement. (Though it doesn't quite work for the most negative value)
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* Mainly:
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* - invert all the bits
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* - add one to the result
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*
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* Of course, this doesn't quite work in Javascript. Take for example the value
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* of -128. This could be represented in 16 bits (big-endian) as 0xff80. But of
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* course, Javascript will do the following:
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*
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* > ~0xff80
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* -65409
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*
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* Whoh there, Javascript, that's not quite right. But wait, according to
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* Javascript that's perfectly correct. When Javascript ends up seeing the
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* constant 0xff80, it has no notion that it is actually a signed number. It
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* assumes that we've input the unsigned value 0xff80. Thus, when it does the
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* binary negation, it casts it into a signed value, (positive 0xff80). Then
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* when you perform binary negation on that, it turns it into a negative number.
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*
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* Instead, we're going to have to use the following general formula, that works
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* in a rather Javascript friendly way. I'm glad we don't support this kind of
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* weird numbering scheme in the kernel.
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*
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* (BIT-MAX - (unsigned)val + 1) * -1
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*
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* The astute observer, may think that this doesn't make sense for 8-bit numbers
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* (really it isn't necessary for them). However, when you get 16-bit numbers,
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* you do. Let's go back to our prior example and see how this will look:
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*
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* (0xffff - 0xff80 + 1) * -1
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* (0x007f + 1) * -1
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* (0x0080) * -1
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*
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* Doing it this way ends up allowing us to treat it appropriately in
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* Javascript. Sigh, that's really quite ugly for what should just be a few bit
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* shifts, ~ and &.
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*/
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/*
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* Endianness doesn't matter for 8-bit signed values. We could in fact optimize
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* this case because the more traditional methods work, but for consistency,
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* we'll keep doing this the same way.
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*/
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function rsint8(buffer, endian, offset)
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{
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var neg;
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neg = buffer[offset] & 0x80;
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if (!neg)
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return (buffer[offset]);
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return ((0xff - buffer[offset] + 1) * -1);
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}
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/*
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* The 16-bit version requires a bit more effort. In this case, we can leverage
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* our unsigned code to generate the value we want to return.
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*/
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function rsint16(buffer, endian, offset)
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{
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var neg, val;
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val = ruint16(buffer, endian, offset);
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neg = val & 0x8000;
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if (!neg)
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return (val);
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return ((0xffff - val + 1) * -1);
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}
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/*
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* We really shouldn't leverage our 32-bit code here and instead utilize the
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* fact that we know that since these are signed numbers, we can do all the
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* shifting and binary anding to generate the 32-bit number. But, for
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* consistency we'll do the same. If we want to do otherwise, we should instead
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* make the 32 bit unsigned code do the optimization. But as long as there
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* aren't floats secretly under the hood for that, we /should/ be okay.
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*/
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function rsint32(buffer, endian, offset)
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{
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var neg, val;
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val = ruint32(buffer, endian, offset);
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neg = val & 0x80000000;
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if (!neg)
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return (val);
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return ((0xffffffff - val + 1) * -1);
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}
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/*
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* The signed version of this code suffers from all of the same problems of the
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* other 64 bit version.
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*/
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function rsint64(buffer, endian, offset)
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{
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var neg, val;
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val = ruint64(buffer, endian, offset);
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neg = val[0] & 0x80000000;
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if (!neg)
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return (val);
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val[0] = (0xffffffff - val[0]) * -1;
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val[1] = (0xffffffff - val[1] + 1) * -1;
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return (val);
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}
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/*
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* We now move onto IEEE 754: The traditional form for floating point numbers
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* and what is secretly hiding at the heart of everything in this. I really hope
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* that someone is actually using this, as otherwise, this effort is probably
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* going to be more wasted.
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*
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* One might be tempted to use parseFloat here, but that wouldn't work at all
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* for several reasons. Mostly due to the way floats actually work, and
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* parseFloat only actually works in base 10. I don't see base 10 anywhere near
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* this file.
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*
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* In this case we'll implement the single and double precision versions. The
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* quadruple precision, while probably useful, wouldn't really be accepted by
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* Javascript, so let's not even waste our time.
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*
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* So let's review how this format looks like. A single precision value is 32
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* bits and has three parts:
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* - Sign bit
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* - Exponent (Using bias notation)
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* - Mantissa
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*
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* |s|eeeeeeee|mmmmmmmmmmmmmmmmmmmmmmmmm|
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* 31| 30-23 | 22 - 0 |
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*
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* The exponent is stored in a biased input. The bias in this case 127.
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* Therefore, our exponent is equal to the 8-bit value - 127.
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*
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* By default, a number is normalized in IEEE, that means that the mantissa has
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* an implicit one that we don't see. So really the value stored is 1.m.
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* However, if the exponent is all zeros, then instead we have to shift
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* everything to the right one and there is no more implicit one.
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*
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* Special values:
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* - Positive Infinity:
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* Sign: 0
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* Exponent: All 1s
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* Mantissa: 0
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* - Negative Infinity:
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* Sign: 1
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* Exponent: All 1s
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* Mantissa: 0
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* - NaN:
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* Sign: *
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* Exponent: All 1s
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* Mantissa: non-zero
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* - Zero:
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* Sign: *
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* Exponent: All 0s
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* Mantissa: 0
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*
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* In the case of zero, the sign bit determines whether we get a positive or
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* negative zero. However, since Javascript cannot determine the difference
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* between the two: i.e. -0 == 0, we just always return 0.
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*
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*/
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function rfloat(buffer, endian, offset)
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{
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var bytes = [];
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var sign, exponent, mantissa, val;
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var bias = 127;
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var maxexp = 0xff;
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/* Normalize the bytes to be in endian order */
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if (endian == 'big') {
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bytes[0] = buffer[offset];
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bytes[1] = buffer[offset+1];
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bytes[2] = buffer[offset+2];
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bytes[3] = buffer[offset+3];
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} else {
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bytes[3] = buffer[offset];
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bytes[2] = buffer[offset+1];
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bytes[1] = buffer[offset+2];
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bytes[0] = buffer[offset+3];
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}
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sign = bytes[0] & 0x80;
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exponent = (bytes[0] & 0x7f) << 1;
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exponent |= (bytes[1] & 0x80) >>> 7;
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mantissa = (bytes[1] & 0x7f) << 16;
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mantissa |= bytes[2] << 8;
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mantissa |= bytes[3];
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/* Check for special cases before we do general parsing */
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if (!sign && exponent == maxexp && mantissa === 0)
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return (Number.POSITIVE_INFINITY);
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if (sign && exponent == maxexp && mantissa === 0)
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return (Number.NEGATIVE_INFINITY);
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if (exponent == maxexp && mantissa !== 0)
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return (Number.NaN);
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/*
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* Javascript really doesn't have support for positive or negative zero.
|
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* So we're not going to try and give it to you. That would be just
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* plain weird. Besides -0 == 0.
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*/
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if (exponent === 0 && mantissa === 0)
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return (0);
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/*
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* Now we can deal with the bias and the determine whether the mantissa
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* has the implicit one or not.
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*/
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exponent -= bias;
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if (exponent == -bias) {
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exponent++;
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val = 0;
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} else {
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val = 1;
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}
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val = (val + mantissa * Math.pow(2, -23)) * Math.pow(2, exponent);
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if (sign)
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val *= -1;
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return (val);
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}
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/*
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* Doubles in IEEE 754 are like their brothers except for a few changes and
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* increases in size:
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* - The exponent is now 11 bits
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* - The mantissa is now 52 bits
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* - The bias is now 1023
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*
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* |s|eeeeeeeeeee|mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm|
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* 63| 62 - 52 | 51 - 0 |
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* 63| 62 - 52 | 51 - 0 |
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*
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* While the size has increased a fair amount, we're going to end up keeping the
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* same general formula for calculating the final value. As a reminder, this
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* formula is:
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*
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* (-1)^s * (n + m) * 2^(e-b)
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*
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* Where:
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* s is the sign bit
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* n is (exponent > 0) ? 1 : 0 -- Determines whether we're normalized
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* or not
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* m is the mantissa
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* e is the exponent specified
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* b is the bias for the exponent
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*
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*/
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function rdouble(buffer, endian, offset)
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{
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var bytes = [];
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var sign, exponent, mantissa, val, lowmant;
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var bias = 1023;
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var maxexp = 0x7ff;
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/* Normalize the bytes to be in endian order */
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if (endian == 'big') {
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bytes[0] = buffer[offset];
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bytes[1] = buffer[offset+1];
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bytes[2] = buffer[offset+2];
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bytes[3] = buffer[offset+3];
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bytes[4] = buffer[offset+4];
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bytes[5] = buffer[offset+5];
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bytes[6] = buffer[offset+6];
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bytes[7] = buffer[offset+7];
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} else {
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bytes[7] = buffer[offset];
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bytes[6] = buffer[offset+1];
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bytes[5] = buffer[offset+2];
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bytes[4] = buffer[offset+3];
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bytes[3] = buffer[offset+4];
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bytes[2] = buffer[offset+5];
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bytes[1] = buffer[offset+6];
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bytes[0] = buffer[offset+7];
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}
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/*
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* We can construct the exponent and mantissa the same way as we did in
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* the case of a float, just increase the range of the exponent.
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*/
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sign = bytes[0] & 0x80;
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exponent = (bytes[0] & 0x7f) << 4;
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exponent |= (bytes[1] & 0xf0) >>> 4;
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|
|
/*
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* This is going to be ugly but then again, we're dealing with IEEE 754.
|
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* This could probably be done as a node add on in a few lines of C++,
|
|
* but oh we'll, we've made it this far so let's be native the rest of
|
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* the way...
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*
|
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* What we're going to do is break the mantissa into two parts, the
|
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* lower 24 bits and the upper 28 bits. We'll multiply the upper 28 bits
|
|
* by the appropriate power and then add in the lower 24-bits. Not
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* really that great. It's pretty much a giant kludge to deal with
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* Javascript eccentricities around numbers.
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*/
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lowmant = bytes[7];
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lowmant |= bytes[6] << 8;
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lowmant |= bytes[5] << 16;
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mantissa = bytes[4];
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mantissa |= bytes[3] << 8;
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mantissa |= bytes[2] << 16;
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mantissa |= (bytes[1] & 0x0f) << 24;
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mantissa *= Math.pow(2, 24); /* Equivalent to << 24, but JS compat */
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mantissa += lowmant;
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|
|
|
/* Check for special cases before we do general parsing */
|
|
if (!sign && exponent == maxexp && mantissa === 0)
|
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return (Number.POSITIVE_INFINITY);
|
|
|
|
if (sign && exponent == maxexp && mantissa === 0)
|
|
return (Number.NEGATIVE_INFINITY);
|
|
|
|
if (exponent == maxexp && mantissa !== 0)
|
|
return (Number.NaN);
|
|
|
|
/*
|
|
* Javascript really doesn't have support for positive or negative zero.
|
|
* So we're not going to try and give it to you. That would be just
|
|
* plain weird. Besides -0 == 0.
|
|
*/
|
|
if (exponent === 0 && mantissa === 0)
|
|
return (0);
|
|
|
|
/*
|
|
* Now we can deal with the bias and the determine whether the mantissa
|
|
* has the implicit one or not.
|
|
*/
|
|
exponent -= bias;
|
|
if (exponent == -bias) {
|
|
exponent++;
|
|
val = 0;
|
|
} else {
|
|
val = 1;
|
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}
|
|
|
|
val = (val + mantissa * Math.pow(2, -52)) * Math.pow(2, exponent);
|
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|
|
if (sign)
|
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val *= -1;
|
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|
|
return (val);
|
|
}
|
|
|
|
/*
|
|
* Now that we have gone through the pain of reading the individual types, we're
|
|
* probably going to want some way to write these back. None of this is going to
|
|
* be good. But since we have Javascript numbers this should certainly be more
|
|
* interesting. Though we can constrain this end a little bit more in what is
|
|
* valid. For now, let's go back to our friends the unsigned value.
|
|
*/
|
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|
|
/*
|
|
* Unsigned numbers seem deceptively easy. Here are the general steps and rules
|
|
* that we are going to take:
|
|
* - If the number is negative, throw an Error
|
|
* - Truncate any floating point portion
|
|
* - Take the modulus of the number in our base
|
|
* - Write it out to the buffer in the endian format requested at the offset
|
|
*/
|
|
|
|
/*
|
|
* We have to make sure that the value is a valid integer. This means that it is
|
|
* non-negative. It has no fractional component and that it does not exceed the
|
|
* maximum allowed value.
|
|
*
|
|
* value The number to check for validity
|
|
*
|
|
* max The maximum value
|
|
*/
|
|
function prepuint(value, max)
|
|
{
|
|
if (typeof (value) != 'number')
|
|
throw (new (Error('cannot write a non-number as a number')));
|
|
|
|
if (value < 0)
|
|
throw (new Error('specified a negative value for writing an ' +
|
|
'unsigned value'));
|
|
|
|
if (value > max)
|
|
throw (new Error('value is larger than maximum value for ' +
|
|
'type'));
|
|
|
|
if (Math.floor(value) !== value)
|
|
throw (new Error('value has a fractional component'));
|
|
|
|
return (value);
|
|
}
|
|
|
|
/*
|
|
* 8-bit version, classy. We can ignore endianness which is good.
|
|
*/
|
|
function wuint8(value, endian, buffer, offset)
|
|
{
|
|
var val;
|
|
|
|
val = prepuint(value, 0xff);
|
|
buffer[offset] = val;
|
|
}
|
|
|
|
/*
|
|
* Pretty much the same as the 8-bit version, just this time we need to worry
|
|
* about endian related issues.
|
|
*/
|
|
function wuint16(value, endian, buffer, offset)
|
|
{
|
|
var val;
|
|
|
|
val = prepuint(value, 0xffff);
|
|
if (endian == 'big') {
|
|
buffer[offset] = (val & 0xff00) >>> 8;
|
|
buffer[offset+1] = val & 0x00ff;
|
|
} else {
|
|
buffer[offset+1] = (val & 0xff00) >>> 8;
|
|
buffer[offset] = val & 0x00ff;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* The 32-bit version is going to have to be a little different unfortunately.
|
|
* We can't quite bitshift to get the largest byte, because that would end up
|
|
* getting us caught by the signed values.
|
|
*
|
|
* And yes, we do want to subtract out the lower part by default. This means
|
|
* that when we do the division, it will be treated as a bit shift and we won't
|
|
* end up generating a floating point value. If we did generate a floating point
|
|
* value we'd have to truncate it intelligently, this saves us that problem and
|
|
* may even be somewhat faster under the hood.
|
|
*/
|
|
function wuint32(value, endian, buffer, offset)
|
|
{
|
|
var val;
|
|
|
|
val = prepuint(value, 0xffffffff);
|
|
if (endian == 'big') {
|
|
buffer[offset] = (val - (val & 0x00ffffff)) / Math.pow(2, 24);
|
|
buffer[offset+1] = (val >>> 16) & 0xff;
|
|
buffer[offset+2] = (val >>> 8) & 0xff;
|
|
buffer[offset+3] = val & 0xff;
|
|
} else {
|
|
buffer[offset+3] = (val - (val & 0x00ffffff)) /
|
|
Math.pow(2, 24);
|
|
buffer[offset+2] = (val >>> 16) & 0xff;
|
|
buffer[offset+1] = (val >>> 8) & 0xff;
|
|
buffer[offset] = val & 0xff;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Unlike the other versions, we expect the value to be in the form of two
|
|
* arrays where value[0] << 32 + value[1] would result in the value that we
|
|
* want.
|
|
*/
|
|
function wuint64(value, endian, buffer, offset)
|
|
{
|
|
prepuint(value[0], 0xffffffff);
|
|
prepuint(value[1], 0xffffffff);
|
|
|
|
if (endian == 'big') {
|
|
wuint32(value[0], endian, buffer, offset);
|
|
wuint32(value[1], endian, buffer, offset+4);
|
|
} else {
|
|
wuint32(value[0], endian, buffer, offset+4);
|
|
wuint32(value[1], endian, buffer, offset);
|
|
}
|
|
}
|
|
|
|
/*
|
|
* We now move onto our friends in the signed number category. Unlike unsigned
|
|
* numbers, we're going to have to worry a bit more about how we put values into
|
|
* arrays. Since we are only worrying about signed 32-bit values, we're in
|
|
* slightly better shape. Unfortunately, we really can't do our favorite binary
|
|
* & in this system. It really seems to do the wrong thing. For example:
|
|
*
|
|
* > -32 & 0xff
|
|
* 224
|
|
*
|
|
* What's happening above is really: 0xe0 & 0xff = 0xe0. However, the results of
|
|
* this aren't treated as a signed number. Ultimately a bad thing.
|
|
*
|
|
* What we're going to want to do is basically create the unsigned equivalent of
|
|
* our representation and pass that off to the wuint* functions. To do that
|
|
* we're going to do the following:
|
|
*
|
|
* - if the value is positive
|
|
* we can pass it directly off to the equivalent wuint
|
|
* - if the value is negative
|
|
* we do the following computation:
|
|
* mb + val + 1, where
|
|
* mb is the maximum unsigned value in that byte size
|
|
* val is the Javascript negative integer
|
|
*
|
|
*
|
|
* As a concrete value, take -128. In signed 16 bits this would be 0xff80. If
|
|
* you do out the computations:
|
|
*
|
|
* 0xffff - 128 + 1
|
|
* 0xffff - 127
|
|
* 0xff80
|
|
*
|
|
* You can then encode this value as the signed version. This is really rather
|
|
* hacky, but it should work and get the job done which is our goal here.
|
|
*
|
|
* Thus the overall flow is:
|
|
* - Truncate the floating point part of the number
|
|
* - We don't have to take the modulus, because the unsigned versions will
|
|
* take care of that for us. And we don't have to worry about that
|
|
* potentially causing bad things to happen because of sign extension
|
|
* - Pass it off to the appropriate unsigned version, potentially modifying
|
|
* the negative portions as necessary.
|
|
*/
|
|
|
|
/*
|
|
* A series of checks to make sure we actually have a signed 32-bit number
|
|
*/
|
|
function prepsint(value, max, min)
|
|
{
|
|
if (typeof (value) != 'number')
|
|
throw (new (Error('cannot write a non-number as a number')));
|
|
|
|
if (value > max)
|
|
throw (new Error('value larger than maximum allowed value'));
|
|
|
|
if (value < min)
|
|
throw (new Error('value smaller than minimum allowed value'));
|
|
|
|
if (Math.floor(value) !== value)
|
|
throw (new Error('value has a fractional component'));
|
|
|
|
return (value);
|
|
}
|
|
|
|
/*
|
|
* The 8-bit version of the signed value. Overall, fairly straightforward.
|
|
*/
|
|
function wsint8(value, endian, buffer, offset)
|
|
{
|
|
var val;
|
|
|
|
val = prepsint(value, 0x7f, -0xf0);
|
|
if (val >= 0)
|
|
wuint8(val, endian, buffer, offset);
|
|
else
|
|
wuint8(0xff + val + 1, endian, buffer, offset);
|
|
}
|
|
|
|
/*
|
|
* The 16-bit version of the signed value. Also, fairly straightforward.
|
|
*/
|
|
function wsint16(value, endian, buffer, offset)
|
|
{
|
|
var val;
|
|
|
|
val = prepsint(value, 0x7fff, -0xf000);
|
|
if (val >= 0)
|
|
wuint16(val, endian, buffer, offset);
|
|
else
|
|
wuint16(0xffff + val + 1, endian, buffer, offset);
|
|
|
|
}
|
|
|
|
/*
|
|
* We can do this relatively easily by leveraging the code used for 32-bit
|
|
* unsigned code.
|
|
*/
|
|
function wsint32(value, endian, buffer, offset)
|
|
{
|
|
var val;
|
|
|
|
val = prepsint(value, 0x7fffffff, -0xf0000000);
|
|
if (val >= 0)
|
|
wuint32(val, endian, buffer, offset);
|
|
else
|
|
wuint32(0xffffffff + val + 1, endian, buffer, offset);
|
|
}
|
|
|
|
/*
|
|
* The signed 64 bit integer should by in the same format as when received.
|
|
* Mainly it should ensure that the value is an array of two integers where
|
|
* value[0] << 32 + value[1] is the desired number.
|
|
*/
|
|
function wsint64(value, endian, buffer, offset)
|
|
{
|
|
var vals = new Array(2);
|
|
|
|
prepsint(value[0], 0x7fffffff, -0xf0000000);
|
|
prepsint(value[1], 0xffffffff, -0xffffffff);
|
|
|
|
/* Fix negative numbers */
|
|
if (value[0] < 0 || value[1] < 0) {
|
|
vals[0] = 0xffffffff - Math.abs(value[0]);
|
|
vals[1] = 0x100000000 - Math.abs(value[1]);
|
|
if (vals[1] == 0x100000000) {
|
|
vals[1] = 0;
|
|
vals[0]++;
|
|
}
|
|
} else {
|
|
vals[0] = value[0];
|
|
vals[1] = value[1];
|
|
}
|
|
|
|
if (endian == 'big') {
|
|
wuint32(vals[0], endian, buffer, offset);
|
|
wuint32(vals[1], endian, buffer, offset+4);
|
|
} else {
|
|
wuint32(vals[0], endian, buffer, offset+4);
|
|
wuint32(vals[1], endian, buffer, offset);
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Now we are moving onto the weirder of these, the float and double. For this
|
|
* we're going to just have to do something that's pretty weird. First off, we
|
|
* have no way to get at the underlying float representation, at least not
|
|
* easily. But that doesn't mean we can't figure it out, we just have to use our
|
|
* heads.
|
|
*
|
|
* One might propose to use Number.toString(2). Of course, this is not really
|
|
* that good, because the ECMAScript 262 v3 Standard says the following Section
|
|
* 15.7.4.2-Number.prototype.toString (radix):
|
|
*
|
|
* If radix is an integer from 2 to 36, but not 10, the result is a string, the
|
|
* choice of which is implementation-dependent.
|
|
*
|
|
* Well that doesn't really help us one bit now does it? We could use the
|
|
* standard base 10 version of the string, but that's just going to create more
|
|
* errors as we end up trying to convert it back to a binary value. So, really
|
|
* this just means we have to be non-lazy and parse the structure intelligently.
|
|
*
|
|
* First off, we can do the basic checks: NaN, positive and negative infinity.
|
|
*
|
|
* Now that those are done we can work backwards to generate the mantissa and
|
|
* exponent.
|
|
*
|
|
* The first thing we need to do is determine the sign bit, easy to do, check
|
|
* whether the value is less than 0. And convert the number to its absolute
|
|
* value representation. Next, we need to determine if the value is less than
|
|
* one or greater than or equal to one and from there determine what power was
|
|
* used to get there. What follows is now specific to floats, though the general
|
|
* ideas behind this will hold for doubles as well, but the exact numbers
|
|
* involved will change.
|
|
*
|
|
* Once we have that power we can determine the exponent and the mantissa. Call
|
|
* the value that has the number of bits to reach the power ebits. In the
|
|
* general case they have the following values:
|
|
*
|
|
* exponent 127 + ebits
|
|
* mantissa value * 2^(23 - ebits) & 0x7fffff
|
|
*
|
|
* In the case where the value of ebits is <= -127 we are now in the case where
|
|
* we no longer have normalized numbers. In this case the values take on the
|
|
* following values:
|
|
*
|
|
* exponent 0
|
|
* mantissa value * 2^149 & 0x7fffff
|
|
*
|
|
* Once we have the values for the sign, mantissa, and exponent. We reconstruct
|
|
* the four bytes as follows:
|
|
*
|
|
* byte0 sign bit and seven most significant bits from the exp
|
|
* sign << 7 | (exponent & 0xfe) >>> 1
|
|
*
|
|
* byte1 lsb from the exponent and 7 top bits from the mantissa
|
|
* (exponent & 0x01) << 7 | (mantissa & 0x7f0000) >>> 16
|
|
*
|
|
* byte2 bits 8-15 (zero indexing) from mantissa
|
|
* mantissa & 0xff00 >> 8
|
|
*
|
|
* byte3 bits 0-7 from mantissa
|
|
* mantissa & 0xff
|
|
*
|
|
* Once we have this we have to assign them into the buffer in proper endian
|
|
* order.
|
|
*/
|
|
|
|
/*
|
|
* Compute the log base 2 of the value. Now, someone who remembers basic
|
|
* properties of logarithms will point out that we could use the change of base
|
|
* formula for logs, and in fact that would be astute, because that's what we'll
|
|
* do for now. It feels cleaner, albeit it may be less efficient than just
|
|
* iterating and dividing by 2. We may want to come back and revisit that some
|
|
* day.
|
|
*/
|
|
function log2(value)
|
|
{
|
|
return (Math.log(value) / Math.log(2));
|
|
}
|
|
|
|
/*
|
|
* Helper to determine the exponent of the number we're looking at.
|
|
*/
|
|
function intexp(value)
|
|
{
|
|
return (Math.floor(log2(value)));
|
|
}
|
|
|
|
/*
|
|
* Helper to determine the exponent of the fractional part of the value.
|
|
*/
|
|
function fracexp(value)
|
|
{
|
|
return (Math.floor(log2(value)));
|
|
}
|
|
|
|
function wfloat(value, endian, buffer, offset)
|
|
{
|
|
var sign, exponent, mantissa, ebits;
|
|
var bytes = [];
|
|
|
|
if (isNaN(value)) {
|
|
sign = 0;
|
|
exponent = 0xff;
|
|
mantissa = 23;
|
|
} else if (value == Number.POSITIVE_INFINITY) {
|
|
sign = 0;
|
|
exponent = 0xff;
|
|
mantissa = 0;
|
|
} else if (value == Number.NEGATIVE_INFINITY) {
|
|
sign = 1;
|
|
exponent = 0xff;
|
|
mantissa = 0;
|
|
} else {
|
|
/* Well we have some work to do */
|
|
|
|
/* Thankfully the sign bit is trivial */
|
|
if (value < 0) {
|
|
sign = 1;
|
|
value = Math.abs(value);
|
|
} else {
|
|
sign = 0;
|
|
}
|
|
|
|
/* Use the correct function to determine number of bits */
|
|
if (value < 1)
|
|
ebits = fracexp(value);
|
|
else
|
|
ebits = intexp(value);
|
|
|
|
/* Time to deal with the issues surrounding normalization */
|
|
if (ebits <= -127) {
|
|
exponent = 0;
|
|
mantissa = (value * Math.pow(2, 149)) & 0x7fffff;
|
|
} else {
|
|
exponent = 127 + ebits;
|
|
mantissa = value * Math.pow(2, 23 - ebits);
|
|
mantissa &= 0x7fffff;
|
|
}
|
|
}
|
|
|
|
bytes[0] = sign << 7 | (exponent & 0xfe) >>> 1;
|
|
bytes[1] = (exponent & 0x01) << 7 | (mantissa & 0x7f0000) >>> 16;
|
|
bytes[2] = (mantissa & 0x00ff00) >>> 8;
|
|
bytes[3] = mantissa & 0x0000ff;
|
|
|
|
if (endian == 'big') {
|
|
buffer[offset] = bytes[0];
|
|
buffer[offset+1] = bytes[1];
|
|
buffer[offset+2] = bytes[2];
|
|
buffer[offset+3] = bytes[3];
|
|
} else {
|
|
buffer[offset] = bytes[3];
|
|
buffer[offset+1] = bytes[2];
|
|
buffer[offset+2] = bytes[1];
|
|
buffer[offset+3] = bytes[0];
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Now we move onto doubles. Doubles are similar to floats in pretty much all
|
|
* ways except that the processing isn't quite as straightforward because we
|
|
* can't always use shifting, i.e. we have > 32 bit values.
|
|
*
|
|
* We're going to proceed in an identical fashion to floats and utilize the same
|
|
* helper functions. All that really is changing are the specific values that we
|
|
* use to do the calculations. Thus, to review we have to do the following.
|
|
*
|
|
* First get the sign bit and convert the value to its absolute value
|
|
* representation. Next, we determine the number of bits that we used to get to
|
|
* the value, branching whether the value is greater than or less than 1. Once
|
|
* we have that value which we will again call ebits, we have to do the
|
|
* following in the general case:
|
|
*
|
|
* exponent 1023 + ebits
|
|
* mantissa [value * 2^(52 - ebits)] % 2^52
|
|
*
|
|
* In the case where the value of ebits <= -1023 we no longer use normalized
|
|
* numbers, thus like with floats we have to do slightly different processing:
|
|
*
|
|
* exponent 0
|
|
* mantissa [value * 2^1074] % 2^52
|
|
*
|
|
* Once we have determined the sign, exponent and mantissa we can construct the
|
|
* bytes as follows:
|
|
*
|
|
* byte0 sign bit and seven most significant bits form the exp
|
|
* sign << 7 | (exponent & 0x7f0) >>> 4
|
|
*
|
|
* byte1 Remaining 4 bits from the exponent and the four most
|
|
* significant bits from the mantissa 48-51
|
|
* (exponent & 0x00f) << 4 | mantissa >>> 48
|
|
*
|
|
* byte2 Bits 40-47 from the mantissa
|
|
* (mantissa >>> 40) & 0xff
|
|
*
|
|
* byte3 Bits 32-39 from the mantissa
|
|
* (mantissa >>> 32) & 0xff
|
|
*
|
|
* byte4 Bits 24-31 from the mantissa
|
|
* (mantissa >>> 24) & 0xff
|
|
*
|
|
* byte5 Bits 16-23 from the Mantissa
|
|
* (mantissa >>> 16) & 0xff
|
|
*
|
|
* byte6 Bits 8-15 from the mantissa
|
|
* (mantissa >>> 8) & 0xff
|
|
*
|
|
* byte7 Bits 0-7 from the mantissa
|
|
* mantissa & 0xff
|
|
*
|
|
* Now we can't quite do the right shifting that we want in bytes 1 - 3, because
|
|
* we'll have extended too far and we'll lose those values when we try and do
|
|
* the shift. Instead we have to use an alternate approach. To try and stay out
|
|
* of floating point, what we'll do is say that mantissa -= bytes[4-7] and then
|
|
* divide by 2^32. Once we've done that we can use binary arithmetic. Oof,
|
|
* that's ugly, but it seems to avoid using floating point (just based on how v8
|
|
* seems to be optimizing for base 2 arithmetic).
|
|
*/
|
|
function wdouble(value, endian, buffer, offset)
|
|
{
|
|
var sign, exponent, mantissa, ebits;
|
|
var bytes = [];
|
|
|
|
if (isNaN(value)) {
|
|
sign = 0;
|
|
exponent = 0x7ff;
|
|
mantissa = 23;
|
|
} else if (value == Number.POSITIVE_INFINITY) {
|
|
sign = 0;
|
|
exponent = 0x7ff;
|
|
mantissa = 0;
|
|
} else if (value == Number.NEGATIVE_INFINITY) {
|
|
sign = 1;
|
|
exponent = 0x7ff;
|
|
mantissa = 0;
|
|
} else {
|
|
/* Well we have some work to do */
|
|
|
|
/* Thankfully the sign bit is trivial */
|
|
if (value < 0) {
|
|
sign = 1;
|
|
value = Math.abs(value);
|
|
} else {
|
|
sign = 0;
|
|
}
|
|
|
|
/* Use the correct function to determine number of bits */
|
|
if (value < 1)
|
|
ebits = fracexp(value);
|
|
else
|
|
ebits = intexp(value);
|
|
|
|
/*
|
|
* This is a total hack to determine a denormalized value.
|
|
* Unfortunately, we sometimes do not get a proper value for
|
|
* ebits, i.e. we lose the values that would get rounded off.
|
|
*
|
|
*
|
|
* The astute observer may wonder why we would be
|
|
* multiplying by two Math.pows rather than just summing
|
|
* them. Well, that's to get around a small bug in the
|
|
* way v8 seems to implement the function. On occasion
|
|
* doing:
|
|
*
|
|
* foo * Math.pow(2, 1023 + 51)
|
|
*
|
|
* Causes us to overflow to infinity, where as doing:
|
|
*
|
|
* foo * Math.pow(2, 1023) * Math.pow(2, 51)
|
|
*
|
|
* Does not cause us to overflow. Go figure.
|
|
*
|
|
*/
|
|
if (value <= 2.225073858507201e-308 || ebits <= -1023) {
|
|
exponent = 0;
|
|
mantissa = value * Math.pow(2, 1023) * Math.pow(2, 51);
|
|
mantissa %= Math.pow(2, 52);
|
|
} else {
|
|
/*
|
|
* We might have gotten fucked by our floating point
|
|
* logarithm magic. This is rather crappy, but that's
|
|
* our luck. If we just had a log base 2 or access to
|
|
* the stupid underlying representation this would have
|
|
* been much easier and we wouldn't have such stupid
|
|
* kludges or hacks.
|
|
*/
|
|
if (ebits > 1023)
|
|
ebits = 1023;
|
|
exponent = 1023 + ebits;
|
|
mantissa = value * Math.pow(2, -ebits);
|
|
mantissa *= Math.pow(2, 52);
|
|
mantissa %= Math.pow(2, 52);
|
|
}
|
|
}
|
|
|
|
/* Fill the bytes in backwards to deal with the size issues */
|
|
bytes[7] = mantissa & 0xff;
|
|
bytes[6] = (mantissa >>> 8) & 0xff;
|
|
bytes[5] = (mantissa >>> 16) & 0xff;
|
|
mantissa = (mantissa - (mantissa & 0xffffff)) / Math.pow(2, 24);
|
|
bytes[4] = mantissa & 0xff;
|
|
bytes[3] = (mantissa >>> 8) & 0xff;
|
|
bytes[2] = (mantissa >>> 16) & 0xff;
|
|
bytes[1] = (exponent & 0x00f) << 4 | mantissa >>> 24;
|
|
bytes[0] = (sign << 7) | (exponent & 0x7f0) >>> 4;
|
|
|
|
if (endian == 'big') {
|
|
buffer[offset] = bytes[0];
|
|
buffer[offset+1] = bytes[1];
|
|
buffer[offset+2] = bytes[2];
|
|
buffer[offset+3] = bytes[3];
|
|
buffer[offset+4] = bytes[4];
|
|
buffer[offset+5] = bytes[5];
|
|
buffer[offset+6] = bytes[6];
|
|
buffer[offset+7] = bytes[7];
|
|
} else {
|
|
buffer[offset+7] = bytes[0];
|
|
buffer[offset+6] = bytes[1];
|
|
buffer[offset+5] = bytes[2];
|
|
buffer[offset+4] = bytes[3];
|
|
buffer[offset+3] = bytes[4];
|
|
buffer[offset+2] = bytes[5];
|
|
buffer[offset+1] = bytes[6];
|
|
buffer[offset] = bytes[7];
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Actually export our work above. One might argue that we shouldn't expose
|
|
* these interfaces and just force people to use the higher level abstractions
|
|
* around this work. However, unlike say other libraries we've come across, this
|
|
* interface has several properties: it makes sense, it's simple, and it's
|
|
* useful.
|
|
*/
|
|
exports.ruint8 = ruint8;
|
|
exports.ruint16 = ruint16;
|
|
exports.ruint32 = ruint32;
|
|
exports.ruint64 = ruint64;
|
|
exports.wuint8 = wuint8;
|
|
exports.wuint16 = wuint16;
|
|
exports.wuint32 = wuint32;
|
|
exports.wuint64 = wuint64;
|
|
|
|
exports.rsint8 = rsint8;
|
|
exports.rsint16 = rsint16;
|
|
exports.rsint32 = rsint32;
|
|
exports.rsint64 = rsint64;
|
|
exports.wsint8 = wsint8;
|
|
exports.wsint16 = wsint16;
|
|
exports.wsint32 = wsint32;
|
|
exports.wsint64 = wsint64;
|
|
|
|
exports.rfloat = rfloat;
|
|
exports.rdouble = rdouble;
|
|
exports.wfloat = wfloat;
|
|
exports.wdouble = wdouble;
|