父进程等待所有子进程完成后再继续

Question

我希望我的父进程等待所有子进程完成后再继续，我有一个解决方案。

int status;

pid_t pid = 0;

int num = 0;


for (int i = 0; i < NUMBER_OF_PROCESSES; i++)
{
    pid = fork();

    if (pid == 0)
    {
        //printf("Hello from Child\n");
        sleep(5 - i);
        printf("Hello from Child %d\n",i + 1);
        num++;
        return 0;
    }
    else if (pid)
    {
        waitpid(pid, &status, 0);
        continue;
    }
    else
    {
        printf("Error\n");
        exit(1);
    }


}

printf("Hello from the process, currentPid : %d, pid : %d\n", getpid(), pid);

return 0;

但似乎我必须等待每个子进程才能完成，有什么办法可以让所有子进程都必须能够运行并行？

Answer 1

使用waitpid

    waitpid(childPid, &returnStatus, 0);  // Parent process waits here for child to terminate.

    if (returnStatus == 0)  // Verify child process terminated without error.  
    {
       std::cout << "The child process terminated normally." << std::endl;    
    }

    if (returnStatus == 1)      
    {
       std::cout << "The child process terminated with an error!." << std::endl;    
    }

parallel.It 中的子进程有可能一个子进程运行很快并在下一个运行秒之前终止，在这种情况下，子进程实际上运行ning连续。

for( int n = 0; n < 4; ++n ) {
    switch( fork()) { 
      /* do stuff, but don't wait() or terminate */
    } 
}

Answer 2

您可以启动所有 child（并保留它们的 pid），然后，您将在循环中使用 waitpid（请参阅等待任何 child 的选项）直到它们没有 child离开了。

听起来不错？

编辑：

#define NB_PROCESSES 5

int main(void)
{
    pid_t pidChild[NB_PROCESSES];
    pid_t stoppedChild;
    int   nbChild                = 0;

    printf("Launching all child.\n");
    for (int i = 0; i < NB_PROCESSES; ++i) {
        if ((pidChild[i] = fork()) == -1) {
            printf("Error while fork the %d child : errno = '%s'.\n", i, strerror(errno));
        } else {
            if (pidChild[i] == 0) {
                sleep(NB_PROCESSES - i);
                printf("Hello from Child %d\n",i);
                return (0);
            } else {
                ++nbChild;
            }
        }
    }

    printf("Waiting all child.\n");
    while (nbChild) {
        stoppedChild = waitpid(WAIT_ANY, NULL, 0);
        for (int i = 0; i < NB_PROCESSES; ++i) {
            if (stoppedChild == pidChild[i]) {
                printf("Child %d stopped.\n", i);
            }
        }
        --nbChild;
    }


    printf("Hello from the process, currentPid : %d\n", getpid());

    return (0);
}

你可以像那样保留他们的 pid。

Answer 3

您只需在循环中启动进程，然后在原始进程中循环等待，直到没有更多生命 child。像这样：

for (int i = 0; i < NUMBER_OF_PROCESSES; i++) {
    pid = fork();   
    if (pid == 0) { // child
        sleep(5 - i);
        printf("Hello from Child %d\n",i + 1);
        num++;
        return 0;
    }
    else if (pid==-1) {
        printf("Error\n");
        break; // out on failure
    }
}
// try to wait for any children while there exists at least one
while ((pid=waitpid(-1,&status,0))!=-1) {
  printf("Process %d terminated\n",pid);
}

因此 children 将同时存在，而 parent 将等待他们的终止。

父进程等待所有子进程完成后再继续

Parent process waits for all child processes to finish before continuing

c

linux

process