Simple recursion failing python

Question

I am baffled by this and I suspect I am making an extremely trivial mistake but here it is anyway:

I want to recursively fill a list with elements. Let's make a simple example. I just want to invert a list in the following manner:

def recurse(x, remaining):

    if len(remaining) == 0:
        assert x == [3, 2, 1] # This does not raise????
        return x
    x.append(remaining.pop())
    recurse(x, remaining)

x = [1, 2, 3]
y = recurse([], x)

However y is None, even when I explicitly check that the value is as expected JUST before returning. What am I missing?

Return values have to be returned, even in recursive functions. — Aran-Fey, Apr 17 '18 at 11:52
Inside of the `if` block, yes. Outside of the `if` block? Not so much. — Aran-Fey, Apr 17 '18 at 11:54
@pilu look at `return recurse(x, remaining)` in @FlyingTeller 's answer — raul.vila, Apr 17 '18 at 11:54

score 2 · Answer 1 · answered Apr 17 '18 at 11:52

2

Only the last call in your recursion does return something, the other calls all don't return anything. What you are missing is handing the return value through to the topmost call of the function like so:

def recurse(x, remaining):

    if len(remaining) == 0:
        assert x == [3, 2, 1] # This does not raise????
        return x
    x.append(remaining.pop())
    return recurse(x, remaining)

x = [1, 2, 3]
y = recurse([], x)

answered Apr 17 '18 at 11:52

FlyingTeller

17,638
3
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53

Also, the `assert` is not throwing any error because the only time it's executed, `x` has exactly that value `[3, 2, 1]` – raul.vila Apr 17 '18 at 11:53

Simple recursion failing python

1 Answers1