调用execv()失败

Question

代码：

static void child() {
    char* args[] = {"/bin/echo", "Hello World!", NULL};
    printf("I'm child! My PID is %d.\n", getpid());
    fflush(stdout);
    execv("/bin/echo", args); // !!
    err(EXIT_FAILURE, "execv() failed");
}

static void parent(__pid_t pid_c) {
    printf("I'm parent! My PID is %d and my child's PID is %d.\n", getpid(), pid_c);
    exit(EXIT_SUCCESS);
}

int main() {
    __pid_t ret;
    ret = fork();

    if (ret == -1) {
        err(EXIT_FAILURE, "fork() failed");
    } else if (ret == 0) {
        child();
    } else {
        parent(ret);
    }

    err(EXIT_FAILURE, "Shouldn't reach here");
}

结果：

I'm parent! My PID is 4543 and my child's PID is 4544.
I'm child! My PID is 4544.

在上面的代码中，我希望将child进程替换为/bin/echo进程，但是echo不能工作。更确切地说，调用execv()失败了。

有什么问题吗？

Answer 1

下列拟议守则：

1. 干净地编译
2. 执行所需的功能
3. 正确地等待子进程完成。
4. 包含所需头文件所需的#include语句。

现在，拟议的守则：

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

static void child() {
    char* args[] = {"/bin/echo", "Hello World!", NULL};
    printf("I'm child! My PID is %d.\n", getpid());
    fflush(stdout);
    execv( args[0], args); 
    err(EXIT_FAILURE, "execv() failed");
}

static void parent(__pid_t pid_c) {
    printf("I'm parent! My PID is %d and my child's PID is %d.\n", getpid(), pid_c);
    wait( NULL );
    exit(EXIT_SUCCESS);
}

int main() {
    __pid_t ret;
    ret = fork();

    if (ret == -1) {
        err(EXIT_FAILURE, "fork() failed");
    } else if (ret == 0) {
        child();
    } else {
        parent(ret);
    }

    err(EXIT_FAILURE, "Shouldn't reach here");
}

由此产生的产出是：

I'm parent! My PID is 31293 and my child's PID is 31294.
I'm child! My PID is 31294.
Hello World!