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What is the output of the following program
#include<stdio.h>
main()
{
int i=0;
fork();
printf("%d",i++);
fork();
printf("%d",i++);
fork();
wait();
}
- 8 Answers
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Answers were Sorted based on User's Feedback
Answer / raghu
totally 8 processes...so each process will print 0 & 1
output: 0101010101010101
| Is This Answer Correct ? | 6 Yes | 1 No |
Answer / sourisengupta
finally we will get 001111.
for the first fork() two 0 will be printed and for secoond
forkk four 1 will be printed.
| Is This Answer Correct ? | 4 Yes | 3 No |
Answer / srikanth
print order cannot be predicted.Total 8 process(including
main) first process prints 0,1
the first child prints 0,1 and the other 2 child/grand child
process print 1's (two 1s). The other four child/grand child
process wont print anything (created after printf)
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / vignesh1988i
for first printf it will print 0, for second printf it will
print 1.. and finally i will have a value 2.
| Is This Answer Correct ? | 5 Yes | 5 No |
Switch (i) i=1; case 1 i++; case 2 ++i; break; case 3 --i; Output of i after executing the program
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