c - fork() 的行为在此示例中不明确

Question

#include<stdio.h>

#include<stdlib.h>

#include<errno.h>
#include<unistd.h>
#include<sys/types.h>
#include<sys/wait.h>

int main(void)

{

pid_t pid;
switch(pid =fork())

 {


case -1:

perror("fork"); /* something went wrong */
exit(1);
/* parent exits */


case 0:
printf(" This is the child process mention in case 0 of switch!, but why case 0 reached?\n");


default:
printf("default case: pid returned by fork is %d\n",pid );

 }
return 0;
}

输出

root@heena:/home/heena/fork# ./a.out 
default case: pid returned by fork is 4640
root@heena:/home/heena/fork#  This is the child process mention in case 0 of switch!, but why case 0 reached?
default case: pid returned by fork is 0

之后我必须按回车键才能进入提示。

为什么案例 0 在默认情况下有效。还有命令提示如何第二次自动出现并执行案例0？然后为什么最后又出现默认值？

Answer 1

这与 fork 关系不大，而与您的 switch 语句有很大关系。

您的子进程从 fork 返回一个 0 返回码，因此它进入 case 0。不幸的是，由于在 case 0 末尾没有 break 语句，因此它会继续执行到 default。这是因为 case 结束时的默认行为只是跳转到下一个 case(如果有的话)。

这与以下没有什么不同:

int i = 1;
switch (i) {
    case 1:  puts ("It was one");
    case 2:  puts ("It was two");
    case 3:  puts ("It was three");
    default: puts ("It was something else");
}

这将以看似精神病的输出结束:

It was one
It was two
It was three
It was something else

要修复它(您的代码，而不是我上面的示例)，请使用:

switch (pid =fork()) {
    case -1:
        perror ("fork");
        exit (1);  // will exit rather than fall through.
    case 0:
        printf (" This is the child process\n");
        break;     // will end switch rather than fall through.
    default:
        printf ("default case: pid returned by fork is %d\n",pid );
                   // nothing needed here since switch is ending anyway.
 }