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fork() 怎么可能返回两个值？

发布于 2024-10-03 04:24:41 字数 64 浏览 4 评论 0原文

既然 C 中的函数始终只返回一个值，那么同样是函数的 fork() 怎么会返回两个值呢？

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鲸落 2024-10-10 04:24:41

fork 函数向创建的子进程返回 0，并向父进程返回子进程的 ID。

这两个单独的进程各自返回一个值。

因此，可以将其更多地视为在每个线程进程上调用一个返回。

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偏闹i 2024-10-10 04:24:41

正如 Gnostus 所说，fork() 函数不会返回两个值。

它的作用是返回一个值，就像所有函数一样，但它返回两次。

一次在父进程中，一次在子进程中。父进程获取返回给它的子进程 ID，子进程获取 0 - 无效的进程 ID，因此代码可以判断它是子进程。

子进程是一个新进程，运行相同的代码，并且与生成它的父进程位于代码中的相同位置。

int cProcessID;

cProcessID = fork();

if (cProcessID == 0) {
    // Child process
} else {
    // Parent process
}

As Gnostus says, the fork() function does not return two values.

What it does is return a single value, the way all functions do, but it returns twice.

Once within the parent process and once within the child. The parent process gets the child's process ID returned to it, the child gets 0 - an invalid process ID, so the code can tell it is the child.

The child is a new process, running the same code and is at the same place in the code as the parent that spawned it.

int cProcessID;

cProcessID = fork();

if (cProcessID == 0) {
    // Child process
} else {
    // Parent process
}

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霞映澄塘 2024-10-10 04:24:41

如果您阅读、构建并运行以下程序，您应该更好地了解正在发生的事情。

#include <stdio.h>
#include <unistd.h>

int main(void) {
    pid_t fk;

    printf("\tbefore fork my pid = %lu\n", (unsigned long)getpid() );

    fflush(stdout); /* This may keep the above print
                       statement from outputing twice. */

    fk = fork(); /* The OS kernel makes a copy of the current process here */

    printf("fork returned %lu and now my pid = %lu\n",
                         (unsigned long)fk, (unsigned long)getpid() );

    return 0;
}

需要 fflush(stdout) 的原因是，由于该进程是通过 fork 重复的，这意味着 stdio 为 stdout 所做的缓冲也会被重复。第一个打印语句末尾的“\n”可能会使其继续并刷新标准输出，但这并不能保证。

If you read, build, and run the following program you should get a better idea of what is going on.

#include <stdio.h>
#include <unistd.h>

int main(void) {
    pid_t fk;

    printf("\tbefore fork my pid = %lu\n", (unsigned long)getpid() );

    fflush(stdout); /* This may keep the above print
                       statement from outputing twice. */

    fk = fork(); /* The OS kernel makes a copy of the current process here */

    printf("fork returned %lu and now my pid = %lu\n",
                         (unsigned long)fk, (unsigned long)getpid() );

    return 0;
}

The reason that the fflush(stdout) is needed is that since the process is duplicated by fork that means that the buffering done for stdout by stdio is duplicated as well. The "\n" at the end of that first print statement may make it go ahead and flush stdout, but this isn't guaranteed.

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