Question

这是使用fork（）来展示它是如何工作的代码（fork.c）。

gcc --version显示gcc (Ubuntu 4.8.4-2ubuntu1~14.04.3) 4.8.4

#include <stdio.h>
int num = 0;
int main(int argc, char*argv[]){
    int pid;
    pid = fork();       
    if(pid == 0){       /*child*/
        num = 1;
    }else if(pid > 0){  /*parent*/
        num = 2;
    }
    printf("%d", num);
}

然后gcc fork.c -o fork进行编译。它编译时没有任何错误，可执行文件正确运行。但我没有明确包含头文件unistd.h，我还检查了所有递归包含的头文件（gcc -H）

. /usr/include/stdio.h
.. /usr/include/features.h
... /usr/include/x86_64-linux-gnu/sys/cdefs.h
.... /usr/include/x86_64-linux-gnu/bits/wordsize.h
... /usr/include/x86_64-linux-gnu/gnu/stubs.h
.... /usr/include/x86_64-linux-gnu/gnu/stubs-64.h
.. /usr/lib/gcc/x86_64-linux-gnu/4.8/include/stddef.h
.. /usr/include/x86_64-linux-gnu/bits/types.h
... /usr/include/x86_64-linux-gnu/bits/wordsize.h
... /usr/include/x86_64-linux-gnu/bits/typesizes.h
.. /usr/include/libio.h
... /usr/include/_G_config.h
.... /usr/lib/gcc/x86_64-linux-gnu/4.8/include/stddef.h
.... /usr/include/wchar.h
... /usr/lib/gcc/x86_64-linux-gnu/4.8/include/stdarg.h
.. /usr/include/x86_64-linux-gnu/bits/stdio_lim.h
.. /usr/include/x86_64-linux-gnu/bits/sys_errlist.h

然后我在所有文件中都grep但我没有找到任何fork（）声明。

如果从未在任何头文件中声明fork（），它怎么能编译而没有任何错误？或者我错过了什么？

Answer 1

在旧版本的C（C99之前：C89或K＆amp; R C）中，您不需要声明来调用函数。然后，您有责任提供正确的参数数量和类型，并且假定返回值为int类型。编译器不为此案例提供任何正确性检查。

但是，编译器应该给你一个警告。这就是GCC 6.3给出的内容：

main.cpp: In function 'main':
main.cpp:5:11: warning: implicit declaration of function 'fork' [-Wimplicit-function-declaration]
    pid = fork();
          ^~~~

Answer 2

In addition to what Ilya Popov said about declarations, function fork() is included in the standard C library, libc, which is implicitly linked with the object files when you build your project. This inclusion does not guarantee that the function is called correctly, but it guarantees its presence.

fork（）神秘地包括在内

2 个答案: