why does pylint complain about Unnecessary "elif" after "return" (no-else-return)?

Question

why does pylint complain about this code block?

R1705: Unnecessary "elif" after "return" (no-else-return)

def f(a):
    if a == 1:
        return 1
    elif a == 2:
        return 2
    return 3

To prevent the error, I had to create a temporary variable, which feels less pleasant.

def f(a):
    if a == 1:
        b = 1
    elif a == 2:
        b = 2
    else:
        b = 3

    return b

Solution:

def f(a):
    if a == 1:
        return 1
    if a == 2:
        return 2
    return 3

score 16 · Accepted Answer · answered Sep 05 '20 at 16:07

16

The purpose of an else block is to define code that will not be executed if the condition is true, so execution wouldn't continue on to the next block.

However, in your code, the main conditional block has a return statement, meaning execution will leave the function, so there's no need for an else block: all subsequent code after the return will, by definition, not be executed if the condition is true. It's redundant. It can be replaced with a simple if.

answered Sep 05 '20 at 16:07

Avner Shahar-Kashtan

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`elif` is a contraction of "else if". – Avner Shahar-Kashtan Sep 05 '20 at 16:33
3

but the code perfectly fine, and it makes sense, right? is this a limitation/bug of pylint? – zyxue Sep 05 '20 at 16:35
1

The code contains a redundant else statement, and pylint notifies you of it, which is precisely its job. – Avner Shahar-Kashtan Sep 05 '20 at 16:37
I see. What's the recommended way of coding such logic? – zyxue Sep 05 '20 at 16:44
4

Simply replace elif with if - they're semantically identical in this situation without the redundant implicit else. – Avner Shahar-Kashtan Sep 05 '20 at 16:54

why does pylint complain about Unnecessary "elif" after "return" (no-else-return)?

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