管道，当父关闭fd [1]时，孩子将从fd [0]得到什么？

Question

#define STACK_SIZE (1024 * 1024)

static char container_stack[STACK_SIZE];
char* const container_args[] = {
    "/bin/bash",
    NULL
};

int pipefd[2];

...

int container_main(void* arg)
{

    ...

    char ch;
    close(pipefd[1]);
    read(pipefd[0], &ch, 1);

    printf("Get EOF [%d] from parent!\n", ch);

    ...

    execv(container_args[0], container_args);
    printf("Something's wrong!\n");
    return 1;
}

int main()
{
    ...

    pipe(pipefd);

    printf("Parent [%5d] - start a container!\n", getpid());

    int container_pid = clone(container_main, container_stack+STACK_SIZE, 
            CLONE_NEWUTS | CLONE_NEWPID | CLONE_NEWNS | CLONE_NEWUSER | SIGCHLD, NULL);


    ...

    close(pipefd[1]);

    ...
    return 0;
}

So, in parent process, I close(pipefd[1]); , then, if I read from read(pipefd[0], &ch, 1); in child, I'll continue jobs in child process. That make sense.

But when I printed the value of ch , it's 0 , I think it should be -1 , which means EOF.

So can anyone tell me why I read 0 from fd[0] in child instead of EOF?

Answer 1

read() returns 0 on EOF ... it returns -1 for errors, so if the other end of the pipe is closed, you expect read to return 0 and not modify ch at all.

That's my mistake, I expected to print EOF to -1 but read() have already deal with EOF and return 0 .

Thanks Dmitri & William Pursell