Python: question about function on list

Question

Hi, everybody!

I have trouble understanding the output of the following code in Python:

def func(a, L= [])
L.append(a)
return L
print (func(1), func(3))

Why is the output [1,3][1,3] and not [1][1,3]?

Please help.

Answer 1

Each function call changes the default parameter L. After the first call, L will be [1]. The second call changes it to [1, 3]. print() only happens after the function was called twice

Answer 2

Maybe this code helps:

https://code.sololearn.com/cvtllrZS8lrA/?ref=app

f() is called twice from within print (). You'll see that the function first does its job two times in a row and only then are the return values handled (also two times in a row). This is because of the chained function calls in print(). It's a bit like a recursive function.

Your code does the same. The function is called twice and both function calls change the value of L. After this happened, the value of L is printed. By the time this happens, L has already been changed twice.

Answer 3

Why does print() happen only after the function is called twice? I really don't get this one.

What happens exactly?
Step 1. func (1) -> L =[1]
Step 2. func (3) -> L = [1,3]
I understand the code and the output until this moment. But why is [1,3][1,3] finally printed, I don't understand.
Is there step 3 hidden somewhere?

Answer 4

After both functions in print() are run, you can read that line as print(L , L) and at that point L = [1,3].

Answer 5

Summing up, in this case first both functions are called and L becomes [1,3]. Printing is done only after all functions are run in the print function. Is this correct?