Re: Aggressive load in gcc when accessing escaped pointer?

[Markus Trippelsdorf <markus at trippelsdorf dot de>]

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