为什么 fork 系统调用的奇怪行为?

问题描述

int main(void) {
    printf("abc");
    fork();
    return 0;
}

这段代码的输出是:

abcabc

为什么会打印两次，即使 fork 系统调用在 printf 语句之后?

Why is it printing twice, even when fork system call is after the printf statement?

Ideone 链接

推荐答案

因为 stdout 是缓冲的，通常是行缓冲的.并且在您的程序中，缓冲区仅在 exit 时间或从 main 返回时刷新(当 fork 不这样做时，会发生两次"失败).

Because stdout is buffered, often line-buffered. And in your program the buffer is flushed only at exit time, or when returning from main (and that happens "twice" when fork don't fail).

尝试在 fork(); 之前添加 fflush(NULL);(您几乎应该总是这样做)

Try adding fflush(NULL); before the fork(); (which you should almost always do)

顺便说一句，你应该始终保持fork的结果并处理三种情况:fork 失败，在孩子，在父母.

BTW, you should always keep the result of fork and handle three cases: fork failure, in child, in parent.

所以 fork 表现得像它应该的那样，但是 printf 没有你想象的天真的直接副作用:它是 缓冲 所以真正的输出可能发生在后者.您还将通过将 fork() 替换为 sleep(15)(输出发生在 exit 时间或结束时)来观察缓冲main，所以 在 sleep 之后，并且在 15 秒内您的程序显然不会输出任何内容)

So fork is behaving as it should, but printf has not the naive immediate side-effect you imagine: it is buffering so the real output may happen latter. You'll also observe the buffering by replacing fork() with sleep(15) (the output happening at exit time, or at end of main, so after the sleep, and for 15 seconds your program won't apparently output anything)

您也可以使用 strace(1)了解正在发生的系统调用...

You might also use strace(1) to understanding what system calls are happening...

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