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Fix endianness bug in write_digit2_separated #2699
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I can confirm this PR fixes the issue:
|
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Thanks for the PR!
include/fmt/chrono.h
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#ifndef _WIN32 | ||
# if defined(__GNUC__) && __BYTE_ORDER__ == __ORDER_BIG_ENDIAN__ | ||
digits = __builtin_bswap64(digits); | ||
# else | ||
if (is_big_endian()) { | ||
digits = ((digits & 0xff00000000000000ull) >> 56) | | ||
((digits & 0x00ff000000000000ull) >> 40) | | ||
((digits & 0x0000ff0000000000ull) >> 24) | | ||
((digits & 0x000000ff00000000ull) >> 8) | | ||
((digits & 0x00000000ff000000ull) << 8) | | ||
((digits & 0x0000000000ff0000ull) << 24) | | ||
((digits & 0x000000000000ff00ull) << 40) | | ||
((digits & 0x00000000000000ffull) << 56); | ||
} | ||
# endif | ||
#endif | ||
memcpy(buf, &digits, 8); |
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I suggest using std::reverse_copy
instead.
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Done.
And improve is_big_endian()
with using of predefined macro. This check is compile time now if it possible (to avoid many is_big_endian()
calls on little endian system and on GCC like compilers).
include/fmt/format.h
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#ifdef _WIN32 | ||
constexpr auto is_big_endian() -> bool { return false; } | ||
#elif defined(__BIG_ENDIAN__) | ||
constexpr auto is_big_endian() -> bool { return true; } | ||
#elif defined(__BYTE_ORDER__) && defined(__ORDER_BIG_ENDIAN__) | ||
constexpr auto is_big_endian() -> bool { | ||
return __BYTE_ORDER__ == __ORDER_BIG_ENDIAN__; | ||
} | ||
#else |
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I suggest moving the ifdefs into the body of is_big_endian
to avoid repeating its signature. I don't think we care about constexpr
here, it will be inlined and optimized away.
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Done
include/fmt/chrono.h
Outdated
memcpy(tmp, &digits, 8); | ||
std::reverse_copy(tmp, tmp + 8, buf); |
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why not std::reverse_copy(digits, digits + 8, buf)
?
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digits
is unsigned long long
.
Without memcpy into temp buffer this is UB.
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Ah, indeed. Let's at least turn 8 into a constant then.
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Done.
Thank you! |
Fix issue #2698
Need testing on big-endian platform.CC @xvitaly