Consider the following two functions. The output printed when fun1(5) is…

Key idea: trace the mutual recursion and remember that ++n in the second function increments n before the recursive call, affecting subsequent prints.

1. fun1(5) prints 5

2. fun2(3) prints 3

3. fun2 increments n to 4 and calls fun1(4); fun1(4) prints 4

4. fun2(2) prints 2

5. fun2 increments to 3 and calls fun1(3); fun1(3) prints 3

6. fun2(1) prints 1

7. fun2 increments to 2 and calls fun1(2); fun1(2) prints 2

8. fun1(2) calls fun2(0) which returns immediately, then fun1(2) prints 2 (second time)

9. Return to fun2(1): it now prints its n (which was incremented to 2), producing another 2

10. Return to fun1(3): it prints 3

11. Return to fun2(2): it prints its n (which is 3)

12. Return to fun1(4): it prints 4

13. Return to fun2(3): it prints its n (which is 4)

14. Return to fun1(5): it prints 5

Final output: 53423122233445