我为什么要在这里分叉5次以上？

Question

所以我在这里有代码,我期望它严格运行ls -l 5次,但它似乎运行的次数要多得多.我在这做错了什么？我想要运行ls 5次,所以我分叉了5次.也许我不理解等待的概念？我经历了大量的教程,似乎没有人能彻底解决使用fork的多个进程.

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>

int main()
{
    pid_t pidChilds[5];

    int i =0;

    for(i = 0; i<5; i++)
    {
        pid_t cpid = fork();
        if(cpid<0)
            printf("\n FORKED FAILED");
        if(cpid==0)
            printf("FORK SUCCESSFUL");
        pidChilds[i]=cpid;
    }





}

Answer 1

当您在 C 中使用 fork 时，您必须想象进程代码和状态被复制到新进程中，此时它从中断处开始执行。

当你在C中使用exec时，你必须想象如果调用成功，整个进程都会被替换。

这是您的代码，经过重新编写以产生预期的行为。请阅读评论。

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>

int main()
{
    pid_t cpid;
    pid_t pidChildren[5];

    int i;
    for (i = 0; i < 5; i++)
    {
        cpid = fork();
        if (cpid < 0) {
            printf("fork failed\n");
        } else if (cpid == 0) {
            /*  If we arrive here, we are now in a copy of the
                state and code of the parent process. */
            printf("fork successful\n");
            break;
        } else {
            /*  We are still in the parent process. */
            pidChildren[i] = cpid;
        }
    }

    if (cpid == 0) {
        /*  We are in one of the children;
            we don't know which one. */
        char *cmd[] = {"ls", "-l", NULL};
        /*  If execvp is successful, this process will be
            replaced by ls. */
        if (execvp(cmd[0], cmd) < 0) {
            printf("execvp failed\n");
            return -1;
        }
    }

    /* We expect that only the parent arrives here. */
    int exitStatus = 0;
    for (i = 0; i < 5; i++) {
        waitpid(pidChildren[i], &exitStatus, 0);
        printf("Child %d exited with status %d\n", i, exitStatus);
    }

    return 0;
}