Thread: problem with creating multiple child with fork() in C

heres the code i have, the number of child is base on the user input, here i just put 5, my intention is to create multiple child to process individual data and then pipe it back to parent where it will store in an array, and print it out after collecting all data

Code:

pipe(fd);
pid=fork();
for(i=1; i<5; i++){
if(pid<0){
exit(-1);
}
else if (pid==0){
close(fd[0]);
//process data and write() to parent
exit(1);
}
else {
wait(NULL);
close(fd[1]);
//read() from child then store in integer array[i]
pipe(fd);
pid=fork();
}
}
// this is outside the for loop
for(a=1; a<5;a++){
printf("%d", array[a]);
}

everything works except that the print for-loop executes twice and give me 2 sets of inputs, i tried to put exit() and wait() in parent but no luck, anyone know what my problem is?

thank you in advance

Your problem is not the loop. When you fork a program the child created starts execution right after the fork call. You are basically duplicating the parent. The reason your print statement executes twice is because it executes in the parent and in the child. In order to solve this put it in an if statement to dectect it pid is greater than 0. I believe this will solve your problem.
I think you will also need to put a wait command before this to assure that all child processes complete before executing the print loop.

Code:

if(pid > 0) 
{
   for(a=1; a<5; ++a)
   {
          printf("%d", array[a]);
   }
}