Question

Please show steps and final result.

int funcB(int);

int funcA(int n) {

if (n <= 5)

return 7;

else

return n + funcB(n - 2);

}

int funcB(int n) {

if (n <= 3)

return 5;

else

return n * funcA(n - 2);

}

int main() {

int num = 8;

cout << funcB(num);

return 0;

}

Answer 1

1. Function B is called with value 8.
2. Since 8 is greater than 3 so we multiply 8 with return value from function A, called with value 6.
3. Now in function A since 6 is greater than 5 so we add 6 with return value from function B, called with value 4.
4. Now in function B since 4 is greater than 3 so we multiply 4 with return value from function A, called with value 2.
5. Now in function, A since 2 is less than 5 so we return 7.
6. That returned value is used in step 4, which multiply 7 with 4 to return the value 28 [(4*7) = 28]   
7. That returned value is used in step 3, which add 28 with 6 to return the value 34 [(28+6) = 34]
8. That returned value is used in step 2, which multiply 34 with 8 to return the value 272 [(34*8) = 272].
9. Finally, 272 is output