Error return a function using if-else shorthand in python

Question

I'm trying to use this if-else shorthand inside a function:

def isFull(): # checks wheteher stack is full
    return True if top == (size - 1) else return False

But my IDE highlights the return saying "Expected expression".

What is wrong with the code, and how do I fix it?

Rather than tell us what your IDE thinks about the code, it's better to see what Python thinks of it when it tries executing the code. This should give you a `SyntaxError` which can be [copied and pasted as properly formatted text](https://meta.stackoverflow.com/questions/359146), and properly shows the problem - [please do not upload images of errors when asking a question](//meta.stackoverflow.com/q/285551), and please show actual errors when possible. — Karl Knechtel, Aug 27 '22 at 04:41
As for the problem, it means exactly what you are told: the part after `else` has to be an *expression*, and using `return` makes a *statement*. It is the same problem as if you had written `return 1 + return 2`, instead of `return 1 + 2`. What is written after `else` is **not** "what to do when the condition isn't met", it's "*what the value should be* when the condition isn't met". I closed this as a duplicate of the general reference question for this construction, which Python calls a *conditional expression*. — Karl Knechtel, Aug 27 '22 at 04:41

score 0 · Accepted Answer · answered Aug 27 '22 at 04:29

You don't repeat return inside the conditional expression. The conditional expression contains expressions, return is a statement.

def isFull():
    return True if top == (size - 1) else False

But there's really no reason for the conditional expression at all, just return the value of the condition, since it's True or False.

def isFull():
    return top == size - 1

1 Answers1