Subject: Re: [boost] Boost.Endian comments
From: Beman Dawes (bdawes_at_[hidden])
Date: 2011-09-07 15:30:35
>>> inline void reorder(int64_t source, int64_t& target)
>>> target = ((source << 0x38) & 0xFF00000000000000)
>>> | ((source << 0x28) & 0x00FF000000000000)
>>> | ((source << 0x18) & 0x0000FF0000000000)
>>> | ((source << 0x08) & 0x000000FF00000000)
>>> | ((source >> 0x08) & 0x00000000FF000000)
>>> | ((source >> 0x18) & 0x0000000000FF0000)
>>> | ((source >> 0x28) & 0x000000000000FF00)
>>> | ((source >> 0x38) & 0x00000000000000FF);
>>> would be more efficient?
>> Yes, but if I read the standard correctly, it is relying on undefined behavior.
>> Both the C and C++ standards say:
>> The value of E1 << E2 is E1 left-shifted E2 bit positions; vacated
>> bits are zero-filled. If E1 has an unsigned type, the value of the
>> result is E1 × 2 [to the power] E2, reduced modulo one more than the
>> maximum value representable in the result type. Otherwise, if E1 has a
>> signed type and non-negative value, and E1×2 [to the power] E2 is
>> representable in the result type, then that is the resulting value;
>> otherwise, the behavior is undefined.
>> In practice I don't think this is a problem, but Boost code generally
>> doesn't rely on code that has undefined behavior. Perhaps it would be
>> OK to #ifdef on the architecture, and use shift-and-mask
>> implementations only on architectures that don't do something weird
>> (like cause an interrupt) in the case covered by the last sentence.
> If you use uint64_t instead of int64_t (always the right thing when
> doing bit-twidling), there shouldn't be any UB, right?
Good point! Each case of "source" in the above code could be changed
to "static_cast<uint32_t>(source)", and then the whole expression cast
back to int32_t.
AFAICS, that does eliminate the UB.
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