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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;
}
- 2 Answers
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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 |
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