On 12/22/2010 03:16 PM, Parmenides wrote:
Hi,
According to the gcc's manual, in an extended assembly statement,
if the string "memory" is added to the list of of clobbered registers,
it will cause "gcc to not keep memory values cached in registers
across the assembler instrution and not optimize stores or loads to
that memory". To this end, IMO, after this kind of statement all
values cached in registers should be write into the corresponding
memory locations. But, the following example illustrates some
different thing.
#define barrier() __asm__ __volatile__("": : :"memory")
int main()
{
int i, n;
int s = 0;
scanf("%d",&n);
for (i = 0; i< n; i++)
s += i;
barrier();
printf("%d\n", s);
return 0;
}
# gcc -S -O tst.c
We get the assembly code like this:
leal -8(%ebp), %eax
pushl %eax
pushl $.LC0
call scanf
movl -8(%ebp), %edx<--- cache n into edx
addl $16, %esp
testl %edx, %edx
jg .L2
movl $0, %ecx
jmp .L4
.L2:
movl $0, %eax<--- cache i into eax
movl $0, %ecx<--- cache s into ecx
.L5:
addl %eax, %ecx
incl %eax
cmpl %eax, %edx
jne .L5
.L4:
<--- at this point, I think the barrier() will cause the values cached
in edx, eax and ecx will be write back to n, i, and s, respectively.
But, this is not really the case.
subl $8, %esp
pushl %ecx
pushl $.LC1
call printf
Well, i and s never have their address taken, so they are not forced
into memory, and the memory clobber can't affect them.
Andrew.
