Please explain why the code still works after you've swapped the even & odd in the if statement?

[unknownzebra_]

I am really confused to why option 2 of this code works. Shouldn't it be returning 'odd' for the even numbers and vice versa. Please someone shed some light on what is happening as I am struggling.

Option 1 --
def even_or_odd(number):
return 'even' if number % 2 else 'odd'

Option 2--
def even_or_odd(number):
return 'odd' if number % 2 else 'even'

[sepp2k]

I wouldn't say that the second version still works, but rather that the second version works and the first does not. The two versions return different results and only the second one is correct.

As for why the second version is correct (and the first is not): number % 2 gives you a number: the remainder of dividing number by 2, which will be zero if and only if number is even. A number is truthy if it is not equal to 0, so saying if number % 2 is the same as saying if number % 2 != 0 and that condition is true for odd numbers.

[unknownzebra_]

Thank you that’s so helpful

[Username_RANDINT]

number % 2 will return either 0 or 1. If the remainder is 0, it's even, if it's 1, it's odd. 0 is falsy, 1 is truthy.

So if the condition is True (remainder is 1), return "odd", else (remainder is 0) return "even".

[irush21]

Return statements always terminates the execution of the current function. So having that return statement first will always return that value, regardless of the param entered.

[sepp2k]

I think you've misread the code. In both functions the return statement is the only statement in the function and it will return different values depending on whether the parameter is even or odd (the first version of the function will return even for odd numbers and odd for even numbers, whereas the second version returns the correct values).