如何使得程序所在的内存页不能被cacke?

firkraag

Had anybody had test the 'memory' in the clobbered list? I tested it in my machine by using gcc with -O and it took no effect.
Here is my code:


#include <stdio.h>
void foo(int x)
{
        __asm__ __volatile__ ("add $0x0a,%0"::"c"(x));
        __asm__ __volatile__("": : :"memory");
        printf("%d\n",x);
}
int main(int argc,char* argv[])
{
        foo(10);
        return 0;
}

It seems should output 10,but the result is 20. Gcc still use the ecx instead the variable 'x' in memory.

BTW, i thought the 'memory' in clobbered list did not invalid the cache, it just told the gcc to reload the variables when being used, quite like the 'volatile' key word in C language, other than
invalid the data in caches and registers.

wanyuan111

回复 21# firkraag
firstly,I don't know what you want to get in your function.Uncachable? or something else?
secondly,whatever you want ,I think ,you should put the instruction :__asm__ __volatile__("": : :"memory") at the front of the instruction : __asm__ __volatile__ ("add $0x0a,%0"::"c"(x));   
And then,please read the memrory barrier carefully,and make sure that you have  understood the instruction :__asm__ __volatile__("": : :"memory") .

最后，请说汉语！

   

firkraag

The gcc manual said:

If your assembler instructions access memory in an unpredictable fashion, add ‘memory’ to the list of clobbered registers. This causes GCC to not keep memory values cached in registers across the assembler instruction and not optimize stores or loads to that memory.

What my code doing is to proving that the gcc does not do things  as it claimed. Put 'memory' in the clobbered list would prevent the program from loading value from cpu cache is mistranslated by somebody.It just prevents the compiler from reusing the value stored in the registers when optimization applied(which means reload the value from memory, if the value is cached in the cpu cache, it will load it from the cpu cache of course.)

A memory barrier that only affects the compiler is a software memory barrier.The barrier() macro is the only software memory barrier.This function inserts a software memory barrier that affects the compiler code generation, but it does not affect the hardware's execution of instructions.

I can not type characters under the os i am using now. If i can, i am glad to use Chinese.