Why does “if foo:” follow the branch even if the function foo returns False?

Question

def foo(a):
    print(\"I\'m foo\")
    return False


if foo:
    print(\"OK\")
else:
    print(\"KO\")

I run it and it returns OK. I know, I should h

Answer 1

python if condition statisfies if the value is not equal to any of these

0, None, "", [], {}, False, ()

Here

def foo(a):
    print("I'm foo")
    return False

>>>foo
<function __main__.foo>

That means variable foo pointing to the function.If you call your function foo(arg) then it will return False as you expecting.So

>>>foo("arg")
False

Answer 2

When you write

 if foo:
    print("OK")
else:
    print("KO")

you are actually testing if the function pointer foo. It is defined, so it prints "OK" and it does not call the function. With the parenthesis, you call the foo function and runs its code.

Answer 3

In this case your code is same as

if foo != None:
    print("OK")
else:
    print("KO")

Result is

"OK"

because foo actually exists.