When the function runs multiple times, the list is nested in the default parameter, but the integer is not

Question

The list, x, isn't initialized to the parameter default value each time the function is operated, but remains in the previous state and adds a new value. That's how the function was executed 5 times, so it became [7, 7, 7, 7, 7]. However, the integer y appears to hold the default value. It has been run 5 times, but it continues to output 2. Why does y finally output 2 instead of 6 when the function is executed 5 times?

I've already read this question, but it explains where the list binds to a function. But I wonder why the principle does not apply to integer default parameters.

def foo(x=[],y=1):
    x.append(7)
    y += 1
    return x, y

print(foo())
print(foo())
print(foo())
print(foo())
print(foo())

([7], 2)
([7, 7], 2)
([7, 7, 7], 2)
([7, 7, 7, 7], 2)
([7, 7, 7, 7, 7], 2)

*"I've already read this question ... But I wonder why the principle does not apply to integer default parameters."* Well that means you didn't read it carefully. It was just 5 mins ago and your question was closed with this duplicate and you are asking the same question again... — Asocia, Jan 24 '21 at 18:33
Yes, I apologize for the repeated questions. And it's true that I didn't read the text in detail. Actually, I tried to read it carefully, but the answers felt too difficult because I was a beginner.. But I was so curious that I was sad that the question was closed. I think I was too hasty. — J. SungHoon, Jan 24 '21 at 18:44

flakes · Accepted Answer · 2021-01-24T18:50:32.637

Int's are immutable, lists are not. When you assign a default value to a parameter you are assigning a reference to that value. Performing y += 1 actually creates a new reference and assigns it to y, it does not change the value of the old reference. When you mutate the container you are not changing the reference. Example:

y = 1
x = y

# This will be true as they have the same reference
print(x is y)  #  True

y += 1
# This will be false as the reference to y has been changed
print(x is y)  #  False

y = []
x = y

# This will be true as they have the same reference
print(x is y)  #  True

y.append(1)
# This will still be true, the reference to the container remains the same.
print(x is y)  #  True

When the function runs multiple times, the list is nested in the default parameter, but the integer is not

1 Answers1