如何在fork()中获取execvp的错误?

问题描述

我有以下代码.

我的问题在代码中

     int main() {

            ....

         if ((uproc.pid = fork()) == -1) {
            return -1;
        }

        if (uproc.pid == 0) {
            /* child */

            const char *argv[3];
            int i = 0;
            argv[i++] = "/bin/sh";
            argv[i++] =  "/my/script.sh";
            argv[i++] = NULL;

            execvp(argv[0], (char **) argv);
            exit(ESRCH);

        } else if (uproc.pid < 0)
            return -1;

        /* parent */
        int status;
        while (wait(&status) != uproc.pid) {
            DD(DEBUG,"waiting for child to exit");
        }

           // If /my/script.sh exit accidentally in some place with error.
           // can I catch this error right here?
          ......
    }

推荐答案

子项的退出状态由wait函数在status变量中提供.

The exit status of the child is provided by the wait function, in the status variable.

您可以使用WEXITSTATUS宏获取退出状态，但前提是程序正常退出(即称为exit或从其main函数返回的程序):

You get the exit status by using the WEXITSTATUS macro, but only if the program exited normally (i.e. called exit or returned from its main function):

if (WIFEXITED(status))
    printf("Child exit status: %d\n", WEXITSTATUS(status));
else
    printf("Child exited abnormally\n");

阅读 wait 手册页以了解更多信息.

Read the manual page for wait for more information.

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