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I'm trying to make a homework assignment where I have to use fork() but I don't know why I can't stop my forks after running them through my for loop:
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
int main(int argc, char *argv[]){
int limit = argc/2;
if(argc%2 == 0){
perror("the number of arguments given must pe even!");
exit(1);
}
int i;
for(i=0; i<=limit; i++){
if(fork()==-1){
perror("childes couldn't be created\n");
exit(1);
}
if(fork()==0){
printf("fork: %d \n",i);
exit(1);
}
wait(0);
}
printf("exiting...\n");
return 0;
}
Output:
warzaru@ubuntu:~/OS/UprocH$ ./exe a b c d
fork: 0
fork: 0
fork: 1
fork: 1
fork: 1
fork: 2
fork: 2
fork: 1
fork: 2
exiting...
exiting...
fork: 2
exiting...
exiting...
fork: 2
exiting...
exiting...
exiting...
warzaru@ubuntu:~/OS/UprocH$ fork: 2
fork: 2
fork: 2
exiting...
629
fork() makes a copy of a process. If you do that in a "for" loop, you have a problem. The first time through the loop you invoke fork() and now you have two processes. Unless you have code to prevent it, both processes with continue to run that loop.
fork() in C. Fork system call is used for creating a new process, which is called child process, which runs concurrently with the process that makes the fork() call (parent process). After a new child process is created, both processes will execute the next instruction following the fork() system call.
Each invocation of fork() results in two processes, the child and the parent. Thus the first fork results in two processes.
Daniel Fischer forced me to provide an answer.
Change:
if(fork()==-1){} // first fork
if(fork()==0){} // second fork
To:
pid = fork();
if(pid == -1)
{
... //failed
}
else if(pid == 0)
{
... //success
}
Or use a switch statement:
switch(fork())
{
case -1:
... // failed
break;
case 0:
... // success
break;
}
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