pipe call and synchronization

Question

I'm experimenting some problems with this code:

#include <stdio.h>
#include <stdlib.h>

#define SIZE 30
#define Error_(x) { perror(x); exit(1); }
int main(int argc, char *argv[]) {

    char message[SIZE];
    int pid, status, ret, fd[2];

    ret = pipe(fd);
    if(ret == -1) Error_("Pipe creation");

    if((pid = fork()) == -1) Error_("Fork error");

    if(pid == 0){ //child process: reader (child wants to receive data from the parent)
        close(fd[1]); //reader closes unused ch.
        while( read(fd[0], message, SIZE) > 0 )
                printf("Message: %s", message);
        close(fd[0]);
    }
    else{//parent: writer (reads from STDIN, sends data to the child)
        close(fd[0]);
        puts("Tipe some text ('quit to exit')");
        do{
            fgets(message, SIZE, stdin);
            write(fd[1], message, SIZE);
        }while(strcmp(message, "quit\n") != 0);
        close(fd[1]);
        wait(&status);
    }
}

Code works fine but I can't explain why! There is no explicit sync between parent and child processes. If the child-process executes before parent, read must return 0 and the process ends, but for some reason it waits for the parent execution. How do you explain this? Maybe I'm missing something.

(Edited)

Why do you expect read to return 0? You're not setting non-blocking I/O anywhere. — Mat, Feb 09 '13 at 12:39
Errors belong on stderr. Please: `#define Error_(x) { fputs(x, stderr); exit(1); }` or `#define Error_(x) { perror(x); exit(1); }` — William Pursell, Feb 09 '13 at 13:31

score 5 · Answer 1 · edited Feb 09 '13 at 12:48

5

Since you didn't use O_NONBLOCK in pipe2, read is blocking by default. Therefore it waits until data are written into the pipe.

edited Feb 09 '13 at 12:48

P.P

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answered Feb 09 '13 at 12:42

md5

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When `(fd[0], message, SIZE) > 0` in while condition is false? – Fabio Carello Feb 09 '13 at 12:53
It becomes false when the pipe is closed at the other end; if you didn't write anything yet, the reader assumes you may want to write something later. IOW the pipe acts as an implicit sync. – loreb Feb 09 '13 at 13:31
2

The pipe is not at all implicit. It is a very explicit synchronization mechanism. Probably the most common. Probably also the simplest. – William Pursell Feb 09 '13 at 13:33

pipe call and synchronization

1 Answers1