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Re: Question on strict aliasing in C.
- From: Georg-Johann Lay <avr at gjlay dot de>
- To: David Brown <david at westcontrol dot com>
- Cc: gcc-help at gcc dot gnu dot org
- Date: Tue, 05 Jul 2011 16:11:08 +0200
- Subject: Re: Question on strict aliasing in C.
- References: <4E11F0FC.7020601@gjlay.de> <iuugt0$i2i$1@dough.gmane.org>
David Brown wrote:
> On 04/07/2011 18:57, Georg-Johann Lay wrote:
>> Hi,
>>
>> this it not a question on GCC, so I apologize for asking a question on
>> C strict aliasing rules here. As I know that some people reading this
>> list are much more familiar with C standard than I am, allow me to ask
>> that question, anyway.
>>
>> Suppose the following C code that tries to implement the standard
>> copysign function, i.e. copy the sign of y into x and return that
>> value. double be 64 bits wide:
>>
>>
>> #define MASK ((unsigned short) 0x8000)
>>
>> double copysign (double x, double y)
>> {
>> unsigned short * const px = (unsigned short*)(char*)&x + 3;
>> unsigned short * const py = (unsigned short*)(char*)&y + 3;
>>
>> *px = *px& ~MASK | *py& MASK;
>> return x;
>> }
>>
>>
>> While I say that this code is not correct because it breaks C's strict
>> aliasing rules (e.g C89/90, Chapter 6.3; C98/99, Chapter 6.5, Clause
>> 7), some other person very well familiar with the standard claims that
>> is correct and no problem.
>>
>> So I want to reassure me if the code is ok or not.
>>
>> I am aware of gcc's -f[no]-strict-aliasing switch, but that's rather
>> technical detail to cope with such presumably incorrect code.
>>
>> Thanks,
>>
>> Johann
>>
>> BTW, gcc does just what I would expect: it returns x unchanged and
>> that is not an "optimizer bug".
>>
>
> Does it get compiled to an empty function?
>
> Anyway, you are correct that strict aliasing means that this function is
> buggy - the compiler can assume that a "double" such as x or y cannot
> share the same address as a "short" (or "unsigned short"), thus the
> pointer "px" cannot point to x, thus x is not changed by the function,
> and can be returned unchanged.
Yes, that's what the function does: it's for a 32-bit machine and
compiles to
copysign:
mov %e2, %d5, %d4
ret
because the return register differs from input register.
> It would be /far/ better to stick to char pointers -
It's not up to me to change these sources or to fix malicious code by
changing build options. I'm just the guy left with evaluation of some
software package.
Johann