GATE CSE 2019 | Question: 17

Question

The following C program is executed on a Unix/Linux system :

#include<unistd.h>
int main()
{
    int i;
    for(i=0; i<10; i++)
        if(i%2 == 0)
            fork();
    return 0;
}

The total number of child processes created is ________________ .

Comments

HOW 0 is divided by 2 or a even no  is still a mystery right

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fork will execute 5 times (i=0,2,4,6,8). so child processes created are 2^5-1=31

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this is good question

Answer 1

Answer is $31$
Fork is called whenever $i$ is even, so we can re-write the code as

for(i=0; i<10; i=i+2)
            fork();

fork() will be called $5$ times($i=0,2,4,6,8)$

$\therefore$ Total number of process $2^5=32$ 

Total number of child process would be $2^5−1=31$

Comments

A little addition

for(i=0;i<2;i++)

{

      fork();

}

is equivalent to

fork();

fork();

Answer 2

void main()
{
   int i;

   for (i=0;i<3;i++)
   {
      fork();
   }
}

this is an example so

 

fork() will be called 5 times(i=0,2,4,6,8) then 

so answer is 2^5-1=31.

source https://stackoverflow.com/questions/26793402/visually-what-happens-to-fork-in-a-for-loop

Comments

good explanation

Answer 3

Total $5$ times fork is called.
Total number of child process would be $2^5 -1 = 31$

Answer 4

You can visualise the fork() through this method...

Comments

wow! you actually build all those calls

Answer 5

The fork() call is made only for the even values of i in the range 0-9(0,2,4,6,8). 5 times hence 2^5-1 =31

Answer 6

From 0 to 9, 0,2,4,6,8 are the values which will satisfy if condition. so fork will be called 5 times. total number of child processes will be 2^5-1=31