Despite returning a call to n * findFactorialRecursively(n - 1), the solution to the last exercise did not change the output:

Execution context: 4
Execution context: 3
Execution context: 2
Execution context: 1
0

Why is the returned value not equal to the product of n in each execution context? The short answer: we didn’t define a base case. To understand the need for a base case, it’s worth discussing the call stack that Swift creates for:

return n * findFactorialRecursively(of: n - 1)

Swift would create a call stack with the following events:

1. findFactorialRecursively(of: 3) = 3 * findFactorialRecursively(of: 2)
2. findFactorialRecursively(of: 2) = 2 * findFactorialRecursively(of: 1)
3. findFactorialRecursively(of: 1) = 1 * findFactorialRecursively(of: 0)

The return value associated with each function call depends on the value returned by n - 1. Do you remember what our function returns when n is equal to 0?

Because our current implementation returns 0 when n is zero, each product in the above call-stack, starting with event 3, will be multiplied by 0. This leads to a 0 solution for each of the contexts above it.

We can fix this with a base case. When the base case is met (n == 0), the factorial method should return a number other than zero.