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