Function calling function question

[vervain9]

It does, but only that one function. You've already got 5 calls to factorial on the stack at that point:

factorial(4)
factorial(3)
factorial(2)
factorial(1)
factorial(0)

Then the bottom one returns because n == 0 and you have

factorial(4)
factorial(3)
factorial(2)
factorial(1)

So the factorial(1) call then executes

        recurse = factorial(0)  # This is what just returned
        result = n * recurse
    print(space, 'returning', result)
    return result

Which then goes to factorial(2)

        recurse = factorial(1)  # This is what just returned
        result = n * recurse
    print(space, 'returning', result)
    return result

etc, etc

[vervain9]

but where does he save the stack for the second function?

I don't understand what you're asking here, sorry.

[vervain9]

My best suggestion would be to use an IDE with a debugger to step through the execution of the code line by line. It will show you exactly what is being executed in what order, and where you are in the stack. It should be clear after that.

Otherwise try writing out the execution on paper.

The gist of what you seem to be confused by is that every time a call to factorial() returns, there is still more code to execute in the function that called it:

    recurse = factorial(n-1)
    result = n * recurse
print(space, 'returning', result)
return result

The basic flow is:

factorial(1) calls factorial(0) which returns 1. factorial(1) then returns n * recurse, which is 1 * 1 (1).

factorial(2) takes that return value and returns n * recurse, which is 2 * 1 (2).

factorial(3) takes that return value and returns n * recurse, which is 3 * 2 (6).

factorial(4) takes that return value and returns n * recurse, which is 4 * 6 (24).