Strange double-brackets arguments in Python

Question

Hello, I've recently had a challenge with this Python code:

def x (a=0):
def y (b=a):
return a+b
if a == 0:
return y
return y(a)
print(x(2) + x()(3))

The last line has this weird thing: x()(3). This makes something strange happen:
1. The x function treats the empty bracket as absence of the "a" argument and uses the predefined value 0 for it:
x() = x(a)
2. The x function takes 0 to 1 arguments. If you put 0 and 3 in one parentheses as in (0, 3), you will get an error, since that's 2 arguments:
x(0, 3) = x(a, b) –> error
3. The y function treats 3 from the second parentheses as the "b" argument, i.e. the second argument that the x function for some reason didn't treat as such:
x()(3) = x(a)(b)

Hence the question: why do these two functions treat these arguments differently? And what are these double brackets anyway? How do we use them?
If you could send me a link to some page where this thing is explained, I'll be grateful too. I just don't even know how to google it 😅

Answer 1

If a is 0, y() will be returned. y() adds b to a. If b is not provided, it will be a; if it is provided (2nd parentheses), it's whatever we input.

I simplified the function for demonstration and added a comment. I recommend to print out values when you are not sure what they would be.

https://code.sololearn.com/c7BMp0hIBIoG/?ref=app

Answer 2

python

Answer 3

Lisa, Simon Sauter, thank you, guys! Now I know that it's nested functions.