Consider the following code: Assuming no possibility of integer overflow, what will be returned by the following call to foo? foo(foo(3)+foo(4));

(15!)/(2!)
  • 3!+4!
  • (7!)!
  • (3!+4!)!
  • 15

This function computes factorial, except that it leaves out 2 from the series of multiplications.

So, foo(3)+foo(4) = 15.