fork()

Question

main()

{

printf(" * ");

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

fork();

printf(" * ");

}

how many times * is printed??

Answer 1

2ⁿ  +1     

Try drawing a tree to understand how each fork works.

http://www.csl.mtu.edu/cs4411.ck/www/NOTES/process/fork/create.html

Comments

I am getting (2^n)+1 '*' only..

for eg for n 2

expanding the for loop (loop unrolling) it can be written as

(1) process (parent)-->

printf("*" );

fork();-----> creates new process (2) with code-----> fork(); printf("*")----->creates another (3)---->printf("*)

fork()------------> creates process (4) ----->printf("*);

printf("*);

So total * printed are 5 for n=2

In general for n for loops 2^n -1 child process would be created so 2^n-1 time * would be printed by child process + 2 times by parent ...in total 2^n+1 '*' would be printed.

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Just review and let me know. the boxes represent an asterisk.horizontal bar is point where fork happens.

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In the diagram for every black node (parent) you have printed '*' again and again. seems like you are considering printf inside for loop while its outside. I ran code on gcc compiler for n 3 it gives 10 "*" printed.

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Oh yes ! I did misinterpreted bracket. But how did you get 10? It should be 9 then right?

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 n for loops 2^n -1 child process would be created so 2^n-1 time * would be printed by child process + 2 times by parent ...in total 2^n+1 '*' would be printed.

Does it make clear...??

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Yes I got it.But you said gcc gave you 10 as answer?

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oh yes i didn't noticed it should be 9....:p

actually its giving different number of '*' everytime I execute.....

http://ideone.com/2hSoxN  ---> it now printed 16 *

and here https://code.hackerearth.com/7e4258A--->its 14
and I have no explanation for that.....may be its because printf buffers output and thus when child executes printf it gets displayed twice....

but i have no proper explanation for the behaviour....