A question about the Compiler

 

Today I read the @ leekayak mentioned a problem http://weibo.com/1465082730/znOSZzU4v

I try to explain it in a simple example. First, let's look at a simpler code.

#include "stdio.h"#include "stdlib.h"#include <pthread.h>int f = 0;int x=  0;void* t1(void*){        while(f==0)        {                if(x!=0) printf("error");        }        return NULL;};void* t2(void*){        x=1;        f=1;        return NULL;};int main(void){         pthread_t* worker =  (pthread_t*) malloc(2*sizeof( pthread_t));         pthread_create(&worker[0],NULL,t1,NULL);         pthread_create(&worker[1],NULL,t2,NULL);         pthread_join(worker[0],NULL);         pthread_join(worker[1],NULL);         free(worker);        return 0;}

We use-O3 to compile and find the problem. Compiler version: GCC version 4.7.2 20120921 (Red Hat 4.7.2-2) (GCC)

At 400788: The command JMP is sent to itself (_ z2t1pv + 0x18) = 400770 + 18 = 400788, and an endless loop is generated.

This is because the part of the T1 function does not see the expectation that F will be changed. Therefore, the compiler assumes that F will not change, which is equivalent to an infinite loop. In fact, this equivalent is wrong, because F may be modified externally. For example, T2

 

Therefore, in this case, adding volatile to the definition of F is a concise method, volatile int f = 0; declares that the compiler cannot assume what the f value will be, but must read it.

 

The other method is to add memory barrier to the display and force the compiler to read it.

While (F = 0) {If (X! = 0) printf ("error ");}

Change

For (_ sync_synchronize (); F = 0 ;__ sync_synchronize ())

{

If (X! = 0) printf ("error ");

}