为什么这里的孩子处理不打印任何东西？

Question

假设之前已经声明了所有变量......因为它们已经存在.子进程不打印任何让我觉得它没有被执行的东西.父进程运行正常,尽管它没有获得共享内存.我为这段代码的长度道歉...

// create 5 child process
for(int k=0;k<5;k++){

    // fork a child process
    pid = fork();

    // error occured on fork
    if (pid < 0) {
        fprintf(stderr, "Fork Failed");
        return 1;
    }
    // this is what the child process will run
    else if (pid == 0) {
        //create a shared mem segment
        segment_id = shmget(IPC_PRIVATE, size, S_IRUSR | S_IWUSR);

        //attach the shared memory segment
        shared_memory = (char *) shmat(segment_id, NULL, 0);

        printf("this is child");

        double x = 0;
        double sum = 0;

        // Run process that sums the function
        for(int i=0; i<n; i++){
            // get random number in range of x1-x2
            x = rand()%(x2 - x1 + 1) + x1;
            sum = sum + f(x);
        }

        //write output to the shared memory segment
        sprintf(shared_memory, "%f", sum);
        execlp("/bin/ls", "ls", NULL);

     }

    // this is what the parent process will run
    else {

       //print output from shared memory
        printf("\n*%s", shared_memory);

        //detach shared memory
        shmdt(shared_memory);

        //Here we add the shared memory to the array
        // To add together at the end
        // but since I cant get the memory to share
        // the array can't be implemented

        //remove the shared memory segment
        shmctl(segment_id, IPC_RMID, NULL);

        wait(NULL);
    }
} // End of for statement

Answer 1

C stdout流在内部缓冲数据.可能是你的"这是孩子"的消息被缓冲了,缓冲区没有被execlp刷新,所以它就消失了.fflush(stdout);在printf之后尝试一下.顺便说一下,你也应该在这之前做到这一点fork(),这样孩子和父母都不会尝试在fork之前编写缓冲输出.