edited by
16,891 views
29 votes
29 votes

A process executes the following code

for(i=0; i<n; i++) fork();

The total number of child processes created is

  1. $n$
  2. $2^n-1$
  3. $2^n$
  4. $2^{n+1} - 1$
edited by

3 Answers

Best answer
37 votes
37 votes

Each fork() creates a child which start executing from that point onward. So, number of child processes created will be $2^n - 1$.

At each fork, the number of processes doubles like from $1 - 2- 4 - 8 ... 2^n$. Of these except $1$, all are child processes.

Reference: https://gateoverflow.in/3707/gate2004-it_64

edited by
3 votes
3 votes

in this solution I have explained more deeply what exactly happening, pls upvote if u like the soln :)

1 votes
1 votes
         F0       // There will be 1 child process created by first fork

      /     \

    F1      F1    // There will be 2 child processes created by second fork

   /  \    /  \

 F2   F2  F2   F2  // There will be 4 child processes created by third fork

/ \   / \ / \  / \

 ...............   // and so on

If we sum all levels of above tree for i = 0 to n-1, we get 2n - 1. So there will be 2n – 1 child processes. On the other hand, the total number of process created are (number of child processes)+1.

 

Note:The maximum number of process is 2n and may vary due to fork failures.Also see this post for more details.

Answer:

Related questions

23 votes
23 votes
4 answers
3
32 votes
32 votes
2 answers
4
Kathleen asked Sep 11, 2014
15,051 views
The data blocks of a very large file in the Unix file system are allocated usingcontinuous allocationlinked allocationindexed allocationan extension of indexed allocation...