96 lines
3.7 KiB
Python
96 lines
3.7 KiB
Python
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##############################################################################
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# #
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# Copyright 2006-2019 WebPKI.org (http://webpki.org). #
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# #
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# Licensed under the Apache License, Version 2.0 (the "License"); #
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# you may not use this file except in compliance with the License. #
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# You may obtain a copy of the License at #
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# #
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# https://www.apache.org/licenses/LICENSE-2.0 #
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# #
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# Unless required by applicable law or agreed to in writing, software #
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# distributed under the License is distributed on an "AS IS" BASIS, #
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# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. #
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# See the License for the specific language governing permissions and #
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# limitations under the License. #
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# #
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##############################################################################
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##################################################################
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# Convert a Python double/float into an ES6/V8 compatible string #
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##################################################################
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def convert2Es6Format(value):
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# Convert double/float to str using the native Python formatter
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fvalue = float(value)
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# Zero is a special case. The following line takes "-0" case as well
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if fvalue == 0:
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return '0'
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# The rest of the algorithm works on the textual representation only
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pyDouble = str(fvalue)
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# The following line catches the "inf" and "nan" values returned by str(fvalue)
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if pyDouble.find('n') >= 0:
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raise ValueError("Invalid JSON number: " + pyDouble)
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# Save sign separately, it doesn't have any role in the algorithm
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pySign = ''
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if pyDouble.find('-') == 0:
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pySign = '-'
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pyDouble = pyDouble[1:]
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# Now we should only have valid non-zero values
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pyExpStr = ''
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pyExpVal = 0
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q = pyDouble.find('e')
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if q > 0:
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# Grab the exponent and remove it from the number
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pyExpStr = pyDouble[q:]
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if pyExpStr[2:3] == '0':
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# Supress leading zero on exponents
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pyExpStr = pyExpStr[:2] + pyExpStr[3:]
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pyDouble = pyDouble[0:q]
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pyExpVal = int(pyExpStr[1:])
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# Split number in pyFirst + pyDot + pyLast
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pyFirst = pyDouble
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pyDot = ''
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pyLast = ''
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q = pyDouble.find('.')
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if q > 0:
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pyDot = '.'
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pyFirst = pyDouble[:q]
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pyLast = pyDouble[q + 1:]
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# Now the string is split into: pySign + pyFirst + pyDot + pyLast + pyExpStr
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if pyLast == '0':
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# Always remove trailing .0
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pyDot = ''
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pyLast = ''
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if pyExpVal > 0 and pyExpVal < 21:
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# Integers are shown as is with up to 21 digits
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pyFirst += pyLast
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pyLast = ''
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pyDot = ''
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pyExpStr = ''
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q = pyExpVal - len(pyFirst)
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while q >= 0:
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q -= 1
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pyFirst += '0'
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elif pyExpVal < 0 and pyExpVal > -7:
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# Small numbers are shown as 0.etc with e-6 as lower limit
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pyLast = pyFirst + pyLast
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pyFirst = '0'
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pyDot = '.'
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pyExpStr = ''
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q = pyExpVal
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while q < -1:
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q += 1
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pyLast = '0' + pyLast
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# The resulting sub-strings are concatenated
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return pySign + pyFirst + pyDot + pyLast + pyExpStr
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