Why is this python if statement not equal to true?

Question

I was going over an assignment, and came across something that confused me, as am I not crazy good with python. Here is the code.

def main():


    list = [1,2]
    x = 2

    if (x in list == True):
       print("hi")

    if (x in list):
       print("Why does this proc?")

main()

I believed the output would be both, but the output is only the second if statement. I know that in C, if you had something like

if (x = 6)

That since there is only one '=' that x is now equal to 6. (As its read, if (), x = 6).

Is something similar happening to this python code? Is it checking 'list == true' first, then from there checking about x being in list?

Any insight would be greatly appreciated!

Answer 1

As you can see, yes, your expression requires explicit grouping:

>>> 2 in [1,2] == True
False
>>> (2 in [1,2]) == True
True

Note that, as @tavo and @MorganThrapp mention, the version without parentheses is doing a chained comparison, checking that 2 in [1,2] and then checking that [1,2] == True. The latter is false, so the full expression is also false.

By the way, don't name your variables after built-ins like list, or you won't be able to use those functions easily.

Also, you don't have to compare the result of an expression to True:

>>> 2 in [1,2]
True

Doing so is the equivalent of asking "is 'the cake is ready' a true statement?" as opposed to "is the cake ready?".