如何在C中恰好启动两个子进程？

Question

我想在 C 程序中启动两个子进程。我使用了fork()两次，但是在检查pids 时，它看起来好像启动了四个进程：

#include <stdio.h>
#include <unistd.h>
#include <wait.h>
int main(int argc, char *argv[]) {

    pid_t one = fork();
    pid_t two = fork();

    if (one == 0) {
        printf("child pid = %d\n", getpid());
    }   

    if (two == 0) {
        printf("child pid = %d\n", getpid());
    }   

    if (one != 0 && two != 0) {
        wait(NULL);
        printf("parent pid = %d\n", getpid());
    }

    return 0;
}

输出是：

g@x:dir$ ./twochilds
child pid = 11408
child pid = 11407
child pid = 11409
child pid = 11409
parent pid = 11406

Answer 1

fork()复制进程，它在父进程中的返回值是实际的子进程 pid，在子地址空间中为 0。假设所有调用都fork()成功，让我们看看发生了什么。

pid_t one = fork();

                          (created by parent)  (created by child_1)
     Parent ---  Child_1 ---  Child_2 ---         Child_3
|   one > 0   |  one = 0  |  

pid_t two = fork(); 

|   one > 0   |  one = 0  |   one > 0 |   one = 0
|   two > 0   |  two > 0  |   two = 0 |   two = 0

if (one == 0) { (Will be executed by child_1 and child_3)
    printf("child pid = %d\n", getpid());
}   

if (two == 0) { (Will be executed by child_2 and child_3
    printf("child pid = %d\n", getpid());
}   

if (one != 0 && two != 0) { (Will be executed by parent only)
    wait(NULL);
    printf("parent pid = %d\n", getpid());
}

这就是为什么你打印了 5 行。

如果您只需要 2 个子进程，则可以执行以下操作：

pid_t pid = fork();
if (pid == -1) {
    // handle error
} else if (pid == 0) { 
    // son process 1
} else {
    pid = fork();
    if (pid == -1) { 
        // handle error
    } else if (pid == 0) { 
        // son process 2
    } else {
        // initial parent process
    }
}