trouble with loop counting

Applications Programming
1

HI there,

I am trying to count manually what this code does but I am stuck and I don't learly see the result. The code works and it compiles and runs but I just don't follow the value of var.

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

int main (int argc, char *argv[]){

    int var = 0;
    pid_t p;
    
    while(++var <5){/*as long as it is less than this it will do fork*/
        if((p=fork()) < 0){
        perror("fork error");
        exit(1);
        }else if(p ==0){ /*if it is the child*/
            var--; /*count down*/
        }
        else{
            if(waitpid(p,NULL,0) != p){/*if waipid is not equal p*/
                perror("waitpid error");
                exit(1);/*exit with status signal1*/
            }/*end if*/
        var++;
        } /*end if/else*/
        var++;
    }/*end while*/
    printf("pid=%d, ppid=%d var=%d, p=%d\n", getpid(), getppid(), var, p);
    exit(0);
}/*end main*/

I understand that in the while condition it says that it should not do something more than 4 times in this case it will fork 4 times. Then in the second if statement it says that if p is the child then we count down (var--) but and yet again at right before closing the while statement we use the counter (var++) again does this mean that whatever was added in the beginning to (++var) at the beginning of the loop then gets subtracted (--var) in the if statement and then added again (var++) just before while?. I cannot follow, can someone give me some insight into how counting the loop and what the code does?

BR,
Bluetxxth

2

This code is intentionally obtuse. They modify the same counting variable many times per loop. They allow the child to continue looping after fork() -- meaning, the child itself will call fork() on the next loop! And the child will have its own, independent value of var!

The most effective way I see to tackle this program is brute force. Make a big chart of what changes when. Mentally work your way through the program and fill it out.

Their random indenting makes it impossible to see where anything begins or ends, too; tis should help.

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

int main (int argc, char *argv[]){

    int var = 0;
    pid_t p;
    
    while(++var <5)
    {/*as long as it is less than this it will do fork*/
        if((p=fork()) < 0)
        {
            perror("fork error");
            exit(1);
        }
        else if(p ==0)
        { /*if it is the child*/
            var--; /*count down*/
        }
        else
        {
            if(waitpid(p,NULL,0) != p){/*if waipid is not equal p*/
                perror("waitpid error");
                exit(1);/*exit with status signal1*/
            }/*end if*/

            var++;
        } /*end if/else*/
        var++;
    }/*end while*/

    printf("pid=%d, ppid=%d var=%d, p=%d\n", getpid(), getppid(), var, p);
    exit(0);
}/*end main*/
3

Hi there,

Thnx for your answer, I kind of imagined that it was going to be complicated... :wink: so I am drawing now..

BR,