On 9 May 2008 at 7:36, Dan wrote:
> What is the correct way to allocate memory that is defined as volatile
> so that all standard compilers will not optomise out writes to the
> memory that are never read again?
What compiler are you using? How are you writing to the memory?
Here's a simple program:
#include <stdlib.h>
#include <string.h>
int main(void)
{
char *x = malloc(128);
memset(x, 0, 128);
free(x);
return 0;
}
Here's the start of the code produced by gcc with -O3 optimization
level:
0x080483e0 <main+0>: lea ecx,[esp+0x4]
0x080483e4 <main+4>: and esp,0xfffffff0
0x080483e7 <main+7>: push DWORD PTR [ecx-0x4]
0x080483ea <main+10>: push ebp
0x080483eb <main+11>: mov ebp,esp
0x080483ed <main+13>: sub esp,0x18
0x080483f0 <main+16>: mov DWORD PTR [ebp-0x8],ecx
0x080483f3 <main+19>: mov DWORD PTR [ebp-0x4],ebx
0x080483f6 <main+22>: mov DWORD PTR [esp],0x80
0x080483fd <main+29>: call 0x8048340 <malloc@[EMAIL PROTECTED]
>
0x08048402 <main+34>: mov DWORD PTR [esp+0x8],0x80
0x0804840a <main+42>: mov DWORD PTR [esp+0x4],0x0
0x08048412 <main+50>: mov ebx,eax
0x08048414 <main+52>: mov DWORD PTR [esp],eax
0x08048417 <main+55>: call 0x8048310 <memset@[EMAIL PROTECTED]
>
0x0804841c <main+60>: mov DWORD PTR [esp],ebx
0x0804841f <main+63>: call 0x8048330 <free@[EMAIL PROTECTED]
>
Notice that even on the highest optimization level, gcc doesn't remove
the call to memset, even with no volatile qualifiers around, and even
though the memory is immediately free()d afterwards.
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