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();
}
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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 |
exit () is used to a) exit () terminates the execution of the program itself b) exit () terminates the execution of the loop c) exit () terminates the execution of the block d) none of the above
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