On 2016.03.18 at 22:05 +0800, Cy Cheng wrote:
> Hi,
>
> Please look at this c code:
>
> typedef struct _PB {
> void* data; /* required.*/
> int f1_;
> float f2_;
> } PB;
>
> PB** bar(PB** t);
>
> void qux(PB* c) {
> bar(&c); /* c is escaped because of bar */
> c->f1_ = 0;
> c->f2_ = 0.f;
> }
>
> // gcc-5.2.1 with -fno-strict-aliasing -O2 on x86
> call bar
> movq 8(%rsp), %rax <= load pointer c
> movl $0, 8(%rax)
> movl $0x00000000, 12(%rax)
>
> // intel-icc-13.0.1 with -fno-strict-aliasing -O2 on x86
> call bar(_PB**)
> movq (%rsp), %rdx <= load pointer c
> movl %ecx, 8(%rdx)
> movq (%rsp), %rsi <= load pointer c
> movl %ecx, 12(%rsi)
>
> GCC only load pointer c once, but if I implement bar in such way:
> PB** bar(PB** t) {
> char* ptr = (char*)t;
> *t = (PB*)(ptr-8);
> return t;
> }
>
> Does this volatile C99 standard rule
> (http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1256.pdf):
>
> 6.5.16 Assignment operators
>
> "3. ... The side effect of updating the stored value of the left operand
> shall occur between the previous and the next sequence point."
No. We discussed this on IRC today and Richard Biener pointed out that
bar cannot make c point to &c - 8, because computing that pointer would
be invalid. So c->f1_ cannot clobber c itself.
--
Markus
