f()+f() gives [3,3,3,3] and why not [3,3,3]

Question

def f(x=[ ]):
     x +=[3]
     return x

print(f()+f())
print(f())

Output:

First:

[3,3,3,3]

Second:

[3,3,3]

Answer 1

because of a default parameter on python will be bound only once on program initiation. so for the first call of f() it returns [3] and change array that x refers to [3].

so

print([3]+f())
print(f())

then the second call on f(), since [] has changed to [3], the result will be [3,3] and also the array under a reference will be changed too.

so

print([3,3]+[3,3])
print(f())

similar to the next line. It will return [3,3] + [3]

print([3,3]+[3,3])
print([3,3,3])

That is how you get it.

Answer 2

The reason the output of f()+f() is [3,3,3,3] is because the same object x is added to itself.

Let's break it down:

After the first call to f() -> x=[3].
After the second call to f() -> x=[3,3]

Now f()+f() = x+x = [3,3] + [3,3] = [3,3,3,3]

Maybe a way to demonstrate is to modify your function to print id(x):

def f(x=[ ]):
    x +=[3]
    print(id(x))
    return x
print(f()+f())
#4370444584
#4370444584
#[3, 3, 3, 3]

If you wanted it to output [3,3,3], you'd have to have the function return a copy of x:

def f(x=[ ]):
    x +=[3]
    return [val for val in x]


print(f()+f())
#[3,3,3]