Output will be 4 only. since the argument is not passed by reference so

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;
}

Question Posted / c++ coder

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Answer Posted / 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.

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