I have a simple program, which I want to simulate the condition that I don 't have enough fork capacity, so I limit the fork number when doing pipe tasks.
let the shell-liked pipeline job written in C++:
ls | cat | cat | cat | cat | cat | cat | cat | cat
I have the code, which run pipe()
and fork()
:
#include <errno.h>
#include <fcntl.h>
#include <iostream>
#include <stdio.h>
#include <sys/wait.h>
#include <unistd.h>
const int fork_limit = 3;
int fork_counter = 0;
static void sig_chld_handler(int signo) {
int status;
pid_t pid;
while ((pid = waitpid(-1, &status, WNOHANG)) > 0) {
printf("received SIGCHLD from child process %d\n", pid);
fork_counter -= 1;
fprintf(stdout, "counter --, %d\n", fork_counter);
}
}
int main(int argc, char **argv) {
signal(SIGCHLD, sig_chld_handler);
char **cmds[9];
char *p1_args[] = {"ls", NULL};
char *p2_args[] = {"cat", NULL};
cmds[0] = p1_args;
cmds[1] = p2_args;
cmds[2] = p2_args;
cmds[3] = p2_args;
cmds[4] = p2_args;
cmds[5] = p2_args;
cmds[6] = p2_args;
cmds[7] = p2_args;
cmds[8] = p2_args;
int pipes[16];
pipe(pipes); // sets up 1st pipe
pipe(pipes + 2); // sets up 2nd pipe
pipe(pipes + 4);
pipe(pipes + 6);
pipe(pipes + 8);
pipe(pipes + 10);
pipe(pipes + 12);
pipe(pipes + 14);
pid_t pid;
for (int i = 0; i < 9; i++) {
// === comment this part to run correctly ===
while (fork_limit < fork_counter) {
usleep(10000);
}
// ===
pid = fork();
if (pid == 0) {
fprintf(stdout, "fork p%d\n", i);
// read
if (i != 0) {
if (dup2(pipes[(i - 1) * 2], 0) < 0) {
fprintf(stderr, "dup2 error\n");
exit(EXIT_FAILURE);
}
}
// write
if (i != 8) {
if (dup2(pipes[i * 2 + 1], 1) < 0) {
fprintf(stderr, "dup2 error\n");
exit(EXIT_FAILURE);
}
}
for (int j = 0; j < 16; j++) {
close(pipes[j]);
}
execvp(*cmds[i], cmds[i]);
} else {
fork_counter += 1;
fprintf(stdout, "counter ++, %d \n", fork_counter);
}
}
for (int j = 0; j < 16; j++) {
close(pipes[j]);
}
waitpid(pid, NULL, 0); // wait the last one.
std::cout << "Parent done." << std::endl;
}
the line while (fork_limit < fork_counter)
is what I restrict the child number.
If I remove the while block, the code runs well, but it hangs if I add this.
I suppose the previous children will die, so that fork_counter -= 1
, and the new child can be forked, but the behavior is not and I cannot figure out why.
Result of without the while
.
counter ++, 1
counter ++, 2
fork p0
fork p1
counter ++, 3
fork p2
counter ++, 4
counter ++, 5
fork p3
fork p4
counter ++, 6
fork p5
counter ++, 7
counter ++, 8
fork p6
fork p7
counter ++, 9
fork p8
received SIGCHLD from child process 13316
counter --, 8
Applications
Desktop
Documents
Downloads
Library
Movies
Music
Pictures
received SIGCHLD from child process 13319
counter --, 7
received SIGCHLD from child process 13318
counter --, 6
received SIGCHLD from child process 13317
counter --, 5
received SIGCHLD from child process 13320
counter --, 4
received SIGCHLD from child process 13322
counter --, 3
received SIGCHLD from child process 13321
counter --, 2
received SIGCHLD from child process 13323
counter --, 1
received SIGCHLD from child process 13324
counter --, 0
Parent done.
Result of having the while
, which means I limit the fork number.
counter ++, 1
counter ++, 2
fork p0
fork p1
counter ++, 3
counter ++, 4
fork p2
fork p3
received SIGCHLD from child process 13291
counter --, 3
counter ++, 4
fork p4
(hang)
The main
program perform (in sequence) the following:
The problem is with the timing of 'close all pipes'. Because the main
is waiting for the first children to complete (while (fork_limit < fork_counter)
), before it can finish step #2.
However, the cat
children (e.g., the 1st cat
) can not finish until their input pipe is closed by ALL processes, including he main
, which is waiting for them to finish. Effectively a deadlock.
Consider small modification to the main
process, which will close the pipes to each children, as soon as the children is forked:
if ( fork() ) {
// Children
...
} else {
// Main - close pipes ASAP.
close(pipes[(i-1)*2]) ;
close(pipes[(i-1)*2+1]);
fork_counter += 1;
fprintf(stdout, "counter ++, %d \n", fork_counter);
}
Probably some modification to the pipe closing in the children is also needed.