How to keep child process alive from a single fork() call? - c

Question

I have a program that creates a child process from a fork() call. The child will continuously receive 1-byte integer input from the user. Once an integer has been sent to the child, the child will send the value to the parent using a pipe. When the parent receives the value, it will add it to an array. -1 is sent to end the program. Once the child sends the parent -1, the parent will sum the previous values in the array and, using another pipe, send this sum value to the child in which the child will print it and terminate the program.
As of right now this is my code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
//Gets input from the user
int getInput() {
int val; char temp;
scanf("%hhd", &temp);
val = temp;
return val;
}
//Sums the values of the entered numbers and returns it
int finish(int arr[], int i) {
int sum = 0;
for (int j = 0; j < i; j++) {sum+= arr[j];}
return(sum);
}
int main() {
int fd[2], fd2[2], val = 0, i = 0, sum, final = -9999999;
int arr[1000];
pid_t pidVal;
//Pipe for sending numbers from child to parent
pipe(fd);
//Pipe for sending the final sum from parent to child
pipe(fd2);
//Create parent and child processes
pidVal = fork();
//Used to make it run continously until -1 is pressed
while(1) {
//Child Process
if (pidVal == 0) {
printf("Child Process (should be 0): %d\n", pidVal);
val = getInput();
printf("You typed: %d\n", val);
//Write to parent
close(fd[0]);
write(fd[1], &val, sizeof(val));
//Read if parent sends sum yet
close(fd2[1]);
read(fd2[0], &final, sizeof(final));
//If sum sent from parent, print and terminate
if (final != -9999999) {
printf("%d\n", final);
exit(0);
}
}
//Parent Process
if (pidVal > 0) {
printf("I'm the parent (should be > 0): %d\n", pidVal);
//Read what child sent to the pipe
close(fd[1]);
read(fd[0], &val, sizeof(val));
//If exit value recieved
if (val == -1) {
//Sum the numbers sent
sum = finish(arr, i);
//Close read directory
close(fd2[0]);
//Write the sum to the pipe
write(fd2[1], &sum, sizeof(sum));
}
//Not -1 as input
else {
//Collect input
arr[i] = val;
i++;
}
}
}
}
However the problem is when I try and send more than one number, the program is stuck, as you can see from this sample output:
I'm the parent (should be > 0): 5673
Child Process (should be 0): 0
3 //My Input
You typed: 3
I'm the parent (should be > 0): 5673
1 //My Input
2 //My Input
I noticed how the child process does not seem to execute at the second iteration, yet the parent does, which leads me to believe the child is terminating after the first run. How can I keep this child alive until the user enters -1? More importantly I want to achieve this functionality with only one parent and one child process derived from only a single fork call for the entire program. Is this possible?

Answer 1

In your child section:
while(1) {
if (pidVal == 0) {
printf("Child Process (should be 0): %d\n", pidVal);
val = getInput();
printf("You typed: %d\n", val);
//Write to parent
close(fd[0]);
write(fd[1], &val, sizeof(val));
//Read if parent sends sum yet
close(fd2[1]);
read(fd2[0], &final, sizeof(final));
//If sum sent from parent, print and terminate
if (final != -9999999) {
printf("%d\n", final);
exit(0);
}
}
...
You're reading a single value from the user, sending it to the parent, then waiting for the result from the parent. The parent meanwhile has read the first value from the child and is waiting for another value, so the parent and child are deadlocked waiting for the other to send them something.
You want the child to loop reading values until it gets -1, then wait for the parent.
if (pidVal == 0) {
printf("Child Process (should be 0): %d\n", pidVal);
do {
val = getInput();
printf("You typed: %d\n", val);
//Write to parent
close(fd[0]);
write(fd[1], &val, sizeof(val));
} while (val != -1);
close(fd2[1]);
read(fd2[0], &final, sizeof(final));
//If sum sent from parent, print and terminate
if (final != -9999999) {
printf("%d\n", final);
exit(0);
}
}