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Consider the following code fragment:
int main(void) {
int m = 4;
mystery ( m );
mystery ( m );
printf("%d", m);
return 0;
}
What is the output on the monitor if mystery is defined as
follows ?
void mystery (int m) {
m = m+3;
}
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Answer / c++ coder
Output will be 4 only.
since the argument is not passed by reference so a local
copy of m is used in the function call which is local to
mystery() it will not have any impact on the variable m
which is used in main() function.
| Is This Answer Correct ? | 6 Yes | 1 No |
Answer / rahul darekar
since in c lang we have to define fun first before we use it
but in this program fun mystery() in not defined and still
it is called so it will give error.
| Is This Answer Correct ? | 3 Yes | 1 No |
Read the following program carefully and write the output of the program. Explain each line of code according to given numbering. #include <stdio.h> #include <unistd.h> #include <stdlib.h> #include <errno.h> 1……………… int main (void) { pid_t pid; 2………………………… pid = fork(); 3…………………………. if (pid > 0) { int i; 4………………………… for (i = 0; i < 5; i++) { 5………………… …………… printf(" I AM VU : %d\n", i); 6………………… …………… sleep(1); } exit(0); } 7………………… ……… else if (pid == 0) { int j; for (j = 0; j < 5; j++) { 8……………………………… printf(" I have no child: %d\n", j); sleep(1); } _exit(0); } else { 9………………………………fprintf(stderr, "can't fork, error %d\n", errno); 10……………… … ………… exit (EXIT_FAILURE); } }
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