56 lines
2.0 KiB
Nim
56 lines
2.0 KiB
Nim
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# From: https://github.com/nim-lang/Nim/pull/11067/
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proc parseHex*[T: SomeInteger](s: string, number: var T, start = 0, maxLen = 0): int
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{.inline, noSideEffect.} =
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## Parses a hexadecimal number and stores its value in ``number``.
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##
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## Returns the number of the parsed characters or 0 in case of an error.
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## If error, the value of ``number`` is not changed.
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##
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## If ``maxLen == 0``, the parsing continues until the first non-hex character
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## or to the end of the string. Otherwise, no more than ``maxLen`` characters
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## are parsed starting from the ``start`` position.
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##
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## It does not check for overflow. If the value represented by the string is
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## too big to fit into ``number``, only the value of last fitting characters
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## will be stored in ``number`` without producing an error.
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runnableExamples:
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var num: int
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doAssert parseHex("4E_69_ED", num) == 8
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doAssert num == 5138925
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doAssert parseHex("X", num) == 0
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doAssert parseHex("#ABC", num) == 4
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var num8: int8
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doAssert parseHex("0x_4E_69_ED", num8) == 11
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doAssert num8 == 0xED'i8
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doAssert parseHex("0x_4E_69_ED", num8, 3, 2) == 2
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doAssert num8 == 0x4E'i8
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var num8u: uint8
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doAssert parseHex("0x_4E_69_ED", num8u) == 11
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doAssert num8u == 237
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var num64: int64
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doAssert parseHex("4E69ED4E69ED", num64) == 12
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doAssert num64 == 86216859871725
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var i = start
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var output = T(0)
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var foundDigit = false
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let last = min(s.len, if maxLen == 0: s.len else: i + maxLen)
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if i + 1 < last and s[i] == '0' and (s[i+1] in {'x', 'X'}): inc(i, 2)
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elif i < last and s[i] == '#': inc(i)
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while i < last:
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case s[i]
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of '_': discard
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of '0'..'9':
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output = output shl 4 or T(ord(s[i]) - ord('0'))
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foundDigit = true
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of 'a'..'f':
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output = output shl 4 or T(ord(s[i]) - ord('a') + 10)
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foundDigit = true
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of 'A'..'F':
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output = output shl 4 or T(ord(s[i]) - ord('A') + 10)
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foundDigit = true
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else: break
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inc(i)
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if foundDigit:
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number = output
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result = i - start
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