Python: Function printing boolean return values [duplicate]

Question

I was working in a small problem in python. This is the problem -

Given an array of ints, compute recursively if the array contains a 6. We'll use the convention of considering only the part of the array that begins at the given index. In this way, a recursive call can pass index+1 to move down the array. The initial call will pass in index as 0.

array6([1, 6, 4], 0) → true
array6([1, 4], 0) → false
array6([6], 0) → true

This is the solution which i came up with.

def array6(intList, index):
    if len(intList) < index + 1:
        print(False)
    else:
        if intList[index] == 6:
            print(True)
        else:
            array6(intList, index + 1)

The script seems to work correctly, except when i change print to return statements and call it as : print(array6([1, 6, 4], 0)) it displays "None" in the console. Why is this? It should ideally display True or False - Right? Or am i missing something in the script?

You forgot to return array6 most probably. Could you show us the code with return? — martianwars
– martianwars, Commented Dec 17, 2016 at 18:50

ForceBru · Accepted Answer · 2016-12-17 18:50:23Z

1

When the execution moves to the most inner else statement, no return statement is found, so None is assumed.

In order to solve that, do:

return array6(intList, index + 1)

answered Dec 17, 2016 at 18:50

ForceBru

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Comments

martianwars · Accepted Answer · 2016-12-17 18:50:55Z

1

It seems like you forgot to return the function array6(intList, index+1). It's crucial in recursive functions that all blocks of your if-else-if chain are returning something. You get a None whenever the function is not returning anything.

This is what the whole code will look like - (look at the comment to know where you went wrong)

def array6(intList, index):
    if len(intList) < index + 1:
        return False
    else:
        if intList[index] == 6:
            return True
        else:
            # you forgot the return here
            return array6(intList, index + 1)

answered Dec 17, 2016 at 18:50

martianwars

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Comments

sberry · Accepted Answer · 2016-12-17 18:51:01Z

1

All Python functions return something. If you do not explicitly return then the return value is None. You are not returning anything, but you were close:

def array6(intList, index):
    if len(intList) < index + 1:
        return False
    else:
        if intList[index] == 6:
            return True
        else:
            return array6(intList, index + 1)

Also, you can remove a level of indentation like this

def array6(intList, index):
    if len(intList) < index + 1:
        return False
    if intList[index] == 6:
        return True
    return array6(intList, index + 1)

Any time an if has a return inside it, you can remove the follow else and de-dent as the function will have already returned.

answered Dec 17, 2016 at 18:51

sberry

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