Read system call blocked sharing a pipe - c

Question

I'm new in Unix systems programming and I'm struggling to understand file descriptors and pipes. Let's consider this simple code:
#include <stdlib.h>
#include <unistd.h>
#include <sys/wait.h>
#include <string.h>
int main() {
int fd[2], p;
char *m = "123456789\n", c;
pipe(fd);
p = fork();
if (p == 0) {
// child
while(read(fd[0], &c, 1) > 0) write(1, &c, 1);
}
else {
// parent
write(fd[1], m, strlen(m));
close(fd[1]);
wait(NULL);
}
exit (0);
}
When I compile and run the code, it outputs 123456789 but the process never ends unless I issue ^C. Actually, both processes appear as stopped in htop.
If the child closes fd[1] prior to read() then it seems to work OK but I don't understand why. The fd are shared between both processes and the parent closes fd[1] after writing. Why then the child doesn't get the EOF when reading?
Thank you in advance!

Answer 1

Well, first of all your parent process is waiting for the child to terminate in the wait(2) system call, whyle your child is blocked in the pipe to read(2) for another character. Both processes are blocked... so you need to act externally to take them off. The problem is that the child process doesn't close it's writing descriptor of the pipe (and also the parent doesn't close its reading descriptor of the pipe, but this doesn't affect here) Simply the pipe blocks any reader while at least one such writing descriptor is still open. Only when all writing descriptors are closed, the read returns 0 to the reader.
When you did the fork(2) both pipe descriptors (fd[0] and fd[1]) were dup()ed on the child process, so you have a pipe with two open file descriptors (one in the parent, one in the child) for writing, and two open descriptors (again, one in the parent, one in the child) for reading, so as one writer remains with the pipe open for writing (the child process in this case) the read made by the child still blocks. The kernel cannot detect this as an anomaly, because the child could still write on the pipe if another thread (or a signal handler) should want to.
By the way, I'm going to comment some things you made bad in your code:
first is that you consider only two cases from fork() for the parent, and for the child, but if the fork fails, it will return -1 and you'll have a parent process writing on a pipe with no reading process, so probably it should block (as I say, this is not your case, but it is an error either) You have always to check for errors from system calls, and don't assume your fork() call is never to fail (think that -1 is considered != 0 and so it falls through the parent's code). There's only one system call that you can execute without checking it for errors, and it is close(2) (although there's much controversy on this)
This same happens with read() and write(). A better solution to your problem would be to have used a larger buffer (not just one char, to reduce the number of system calls made by your program and so speed it up) and use the return value of read() as a parameter on the write() call.
Your program should (it does on my system, indeed) work with just inserting the following line:
close(fd[1]);
just before the while loop in the child code, as shown here:
#include <stdlib.h>
#include <unistd.h>
#include <sys/wait.h>
#include <string.h>
int main() {
int fd[2], p;
char *m = "123456789\n", c;
pipe(fd);
p = fork();
if (p == 0) {
// child
close(fd[1]); // <--- this close is fundamental for the pipe to work properly.
while(read(fd[0], &c, 1) > 0) write(1, &c, 1);
}
else if (p > 0) {
// parent
// another close(fd[0]); should be included here
write(fd[1], m, strlen(m));
close(fd[1]);
wait(NULL);
} else {
// include error processing for fork() here
}
exit (0);
}

Answer 2

If the child closes fd[1] prior to read() then it seems to work OK but I don't understand why.
That's what you need to do. There's not much more to it than that. A read from the read end of a pipe won't return 0 (signaling EOF) until the kernel is sure that nothing will ever write to the write end of that pipe again, and as long as it's still open anywhere, including the process doing the reading, it can't be sure of that.