Consider the following code fragment: int main(void) { int m = 4;

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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Consider the following code fragment: int main(void) { int m = 4; mystery ( m ); ..

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

Consider the following code fragment: int main(void) { int m = 4; mystery ( m ); ..

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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