Question

当我尝试在Xcode中构建代码时，我遇到了问题朋友，我是C的noobie，我需要一些帮助

这是我的 main.c 文件

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

int main (int argc, char *argv[])
       {
       pid_t pid;
       int status;

       if (argc < 2) {
           printf ("Something goes wrong!\n");
           return 1;
       }

    for (;;) {
        pid = fork ();
        if (pid == -1) {
            perror ("fork");
            exit (1);
        }
        if (pid == 0) {
            execvp (argv[1], &argv[1]);
        } else {
            wait (&status);
            if (WIFEXITED (status))
            {
                printf ("%s exited with return code %d\n",
                        argv[1], WEXITSTATUS (status));
                if (WEXITSTATUS (status) != 0)
            }
            else if (WIFSIGNALED(status))
            {
                printf ("%s terminated by signal number %d\n",
                        argv[1], WTERMSIG (status));
            }
            /* we want to give time to user for read output of program */
            sleep (1);
        }
    }

    return 0;
}

此行之后

if (WEXITSTATUS (status) != 0)

我有一个解析问题预期声明任何人都可以告诉我为什么错了？谢谢。

Answer 1

只需更改此

printf ("%s exited with return code %d\n",
    argv[1], WEXITSTATUS (status));
if (WEXITSTATUS (status) != 0)

到这个

printf ("%s exited with return code %d\n",
    argv[1], WEXITSTATUS (status));
if (WEXITSTATUS (status) != 0);

或

printf ("%s exited with return code %d\n",
    argv[1], WEXITSTATUS (status));
if (WEXITSTATUS (status) != 0) {}

解析预期声明C程序

1 个答案: