Why doesn't the copy of a list in a function work?

Question

b = [0]

def copyalist(b):
    b = [1, 2, 3]
    print(b)

copyalist(b)
print(b)

The outputs are below:

[1, 2, 3]
[0]

The first line indicates that in the function, b was set to [1, 2, 3]; However, when you print(b) out of the function,the second output says that b is still [0].

I don't understand that, why the outer b is not changed?

I also tried b = copy.deepcopy([1, 2, 3]), the outputs are the same.

However, the following code works well:

b = [0]

def copyalist(b):
    b += [1, 2, 3]
    print(b)

copyalist(b)
print(b)

The outputs are below:

[0, 1, 2, 3]
[0, 1, 2, 3]

Answer 1

In python, lists are passed as function arguments only by reference, ie, only the memory address of the first element is given. When defining a new b inside the function, you just change the position in memory to which the inner variable b refers, but the outer b still points to the original position. Vice versa, when you do b += [1, 2, 3] , you change the content inside the cell referenced by the inner b and, since inner and outer b point to the same cells, it reflects in a change of the outer b as well.

Answer 2

b = copyalist(b) 

and return b in the function

def copyalist(b):
    b = [1, 2, 3]
    return(b)

When you are defining something in a function it is only used inside that function.

Answer 3

This code:

def copyalist(b):
    b = [1, 2, 3]
    print(b)

will only mean to remap the variable name b to a new list, but not modifying the original b . If you mean to modify the original list, you have to explicitly tell Python to replace the actual content. The way to do is:

def copyalist(b):
    b[:] = [1, 2, 3]
    print(b)

Answer 4

This is due to a difference in global and local variables. You can add global to your function to get the desired result.

b = [0]

def copyalist():
    global b
    b = [1, 2, 3]
    print(b)

copyalist()
print(b)

OUTPUT

[1, 2, 3]
[1, 2, 3]

A more in depth summary is here