Jan Kończak

Yet another reason that makes concurrent programming harder.

Below, there is a piece of code that can be compiled by issuing:

gcc -g -O0 sb.c -o sb

The program forks and uses shared memory, so concurrency is involved.
Analyze the code and tell which results can be output.
Remember that newly allocated anonymous shared memory is always zeroed.

Run the program multiple times with the following command, and observe the results:

( for((i=0;i<10000;++i)); do ./sb; done ) | sort | uniq -c

Finally, explain the results.

sb.c

#include <unistd.h>
#include <sys/mman.h>
#include <stdatomic.h>
#include <stdio.h>
 
struct data {
    atomic_char vA, vB, control[3];
    volatile char data[64*3];
};
 
int main(){
    struct data * const s = mmap(NULL, sizeof(struct data),
                                 PROT_READ|PROT_WRITE, MAP_SHARED|MAP_ANONYMOUS, -1, 0);
 
    atomic_char * const control = s->control;
    volatile char * const a = s->data + 64;
    volatile char * const b = s->data + 64*2;
 
    volatile char dummy = s->vA + s->vB + *a + *b;
 
    if(!fork()){
 
        control[0]++;
        while(control[0]!=2);
 
        *b = 1;
        //while(--dummy);
        s->vA = *a;
 
        control[1]=1;
        return 0;
 
    } else {
 
        control[0]++;
        while(control[0]!=2);
 
        *a = 1;
        s->vB = *b;
 
        while(control[1]!=1);
 
        printf("%d %d\n",s->vA, s->vB);
        return 0;
    }
}

Jan Kończak

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Yet another reason that makes concurrent programming harder.

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