First let’s consider a program which loops 2 times:

[sourcecode language=”c”]

#include <stdio.h>

#include <sys/types.h>

#include <unistd.h>

#define N 2

int main(){

int i;

for (i = 1; i <= N; i++) fork();

printf("Hey! from %d\n", getpid());

}

[/sourcecode]

This program prints “Hey”, 4 times to the console, hence it creates 3 child processes. Let’s figure out how!

**Note: The child processes have their local copies of variables, hence any operations on the variables won’t affect the parent’s copy or the other children’s copy.**

1) The parent process (P0) runs through the loop 2 times, hence creating 2 child processes (P1 and P2). These child processes get their local copies of ‘i’ which is 1 for the first child and 2 for the second child. Thus, P2 runs through the loop 1 time while the P1 runs through the loop 2 times.

**Note: The child always start executing from the instruction immediately after fork, which is i++ in our case.**

2) P2 increments i from 2 to 3, followed by the comparison “i <= N” which obviously evaluates to false. Hence, P2 doesn’t create any processes.

3) P1 increments i from 1 to 2, it then executes fork() which creates another child with i as 2, lets call this child “P1_1″. After that it increments i from 2 to 3 and exits out of the loop.

4) P1_1 increments i from 2 to 3, and exits out of the loop.

So, at the end of the loop we have 3 children (P1, P2, P1_1).

Let’s now figure out the number of children for a program which loops ‘N’ times.

Let Pn be the nth child of the parent process, Pn_n be the nth child of the nth child of the parent process and so on.

Children with ‘i’ as 1: 1 (P1)

Children with ‘i’ as 2: 2 (P1_1, P2)

Children with ‘i’ as 3: 4 (P1_2, P2_1, P3, P1_1_1)

Do you notice a pattern in the above written numbers?

Number of children with ‘i’ as 1: 1

Number of children with ‘i’ as 2:

Number of children with ‘i’ as 3:

..

Number of children with ‘i’ as N:

Now we know that sum of G.P is

Hence, total number of child precesses created is: =

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