Re: How to understand the "memory" and the volatile keyword?

[Richard Biener <richard dot guenther at gmail dot com>]

On January 29, 2017 4:56:46 PM GMT+01:00, parmenides via gcc <gcc@gcc.gnu.org> wrote:
>Hi,
>
>   I want to know how to use inline assembler instruction, and wonder 
>what is the meaning of "memory" in clobbered register list. According
>to 
>the manual of GCC, the "memory" will cause GCC to not keep memory
>values 
>cached in registers across the assembler instruction and not optimize 
>stores or loads to that memory. IMO, that means the "memory" notifies 
>GCC the assembler instruction meant to modify memory locations; If 
>variables are cached into registers before the assembler instruction, 
>then these varialbes should be written back to relevant memory
>locations 
>before the assembler instruction, and read when being used again after 
>the assembler instruction. On the other hand, the "memory" affected 
>inputs and outputs of the assembler instruction only. Then the volatile
>
>keyword play its role, if the memory affected is not listed in the 
>inputs and outputs of the assembler instruction. So, I think the 
>"memory" and the volatile keyword should be used to implement a
>barried. 
>I write the following C code to test the effect of the twos:
>
>         /* asm.c */
>         int foobar() {
>           int i, t, sum = 0;
>
>           for (i = 0; i <= 10; i++) {
>             t = 1 << i;
>             sum += t;
>           }
>           asm volatile ("nop" : : : "memory");
>           return sum;
>         }
>         ... ...
>
>I compiled the asm.c using gcc 4.4.7:
>         gcc -S -O1 asm.c
>
>Then, the asm.s is generated as follows:
>
>         foobar:
>             pushl   %ebp
>             movl    %esp, %ebp
>             pushl   %ebx
>             movl    $0, %eax
>             movl    $0, %ecx
>             movl    $1, %edx
>         .L2:
>             movl    %edx, %ebx
>             sall    %cl, %ebx
>             addl    %ebx, %eax
>             addl    $1, %ecx
>             cmpl    $11, %ecx
>             jne     .L2
>         #APP
>         # 10 "hello.c" 1
>             nop
>         # 0 "" 2
>         #NO_APP
>             negl    %eax
>             popl    %ebx
>             popl    %ebp
>             ret
>         ... ...
>
>It is obvious that every variables are cached into register, and even 
>though the 'sum' is used after the barrier, it does not written back to
>
>its memory location and read again. I think my idea about the "memory" 
>and the volatile keyword must be misunderstanding. Any suggestion?

Local variables are not memory unless they have their address taken and it escapes.

Richard.