Tono
Tono

Reputation: 39

parent doesn't create child?

I try to create 4 child processes who will work simultaneously, but the output of my program is quite random: sometimes one of the processes is not created (the printf statement is not executed). I can not understand the reason for this, because when I use the wait() function in my parent it should wait for it's children to finish? Here is my code:

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

#define NR_OF_PROCESSES 4

void readProcess(void);
void tokenProcess(void);
void calculatorProcess(void);
void errorProcess(void);

void (*functionTable[NR_OF_PROCESSES]) (void) = {readProcess, tokenProcess, calculatorProcess, errorProcess};

int main(void) {

    int i;
    int status;
    for (i = 0; i < NR_OF_PROCESSES; i++) {

        int pid = fork();
        if (pid < 0) {
            perror("could not fork");
            return 0;
        }
        if (pid == 0) {
            functionTable[i]();
            exit(0);
        }
    }

    printf("parent is waiting");
    wait(status);

    return (0);
}

void readProcess(void) {
    printf("readprocess running, PID=%d, PPID=%d\n",getpid(),getppid());
}

void tokenProcess(void) {
    printf("tokenprocess running, PID=%d, PPID=%d\n",getpid(),getppid());
}

void calculatorProcess(void) {
    printf("calculatorprocess running, PID=%d, PPID=%d\n",getpid(),getppid());
}

void errorProcess(void) {
    printf("errorprocess running, PID=%d, PPID=%d\n",getpid(),getppid());
}

Also, I want to add interprocess communication with pipes later on. Will this be possible if I implement the processes this way? Any better solution?

Upvotes: 1

Views: 105

Answers (1)

Jonathan Wakely
Jonathan Wakely

Reputation: 171263

You only wait for one child, you probably want to call wait for each child.

Upvotes: 1

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