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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 Posted / 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 |
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