The danger with implicit if statements in Python

Published on 3 November 2014.

In Python we can put an expression in an if statement that is not a boolean. For example:

a_list = [1, 2, 3]
if a_list:
    # do something

The expression will evaluate to either true or false. Some examples of expressions that will evaluate to false:

• [] (empty list)
• "" (empty string)
• 0 (the number 0)
• None (the null value)

Some examples of expressions that will evaluate to true:

• [1, 2] (non-empty list)
• "hello" (non-empty string)
• 88 (the number 88)

So if we are only interested in knowing if a value is truthy, we do not need to make an explicit comparison in the if statement. The above example with an explicit comparison would look like this:

if a_list != []:
    # do something

We can argue that the first example read better because there is less cruft in the expression, but there is one real danger in being implicit. Consider a function that returns either a number or None if no number could be returned. We want to run some code only if we get a number back:

number = give_me_a_number()
if number:
    # do something

This works fine for most numbers:

• 1 -> we do something (expected)
• 2 -> we do something (expected)
• None -> we do not do something (expected)

Except for:

• 0 -> we do not do something (not expected)

The number 0 is a number, so we would like to do something with it. But on the other hand, the number 0 evaluates to false. So with an implicit check, it is not considered truthy, and we will not enter the if block. What we should have done instead was this:

if number is not None:
    # do something

I have made this mistake more than once, and I’m starting to think that explicit if statements should always be used except in special cases.

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