Python 中的可变和不可变

1numbers = [1, 2, 3]
2numbers[0] = 100
3print(numbers)  # [100, 2, 3]

1text = "hello"
2# text[0] = "H"  # TypeError: 'str' object does not support item assignment
3
4text = "H" + text[1:]  # Creates a new string
5print(text)  # "Hello"

 1# Mutable example
 2a = [1, 2, 3]
 3b = a
 4b[0] = 100
 5print(a)  # [100, 2, 3] -> a is also changed
 6
 7# Immutable example
 8x = 10
 9y = x
10y = 20
11print(x)  # 10 -> x is unchanged

 1# Immutable int behavior
 2a = 10
 3print(id(a))  # e.g., 140715920176592
 4a += 1
 5print(id(a))  # e.g., 140715920176624 -> ID has changed
 6
 7# Mutable list behavior
 8b = [1, 2, 3]
 9print(id(b))  # e.g., 2819127951552
10b.append(4)
11print(id(b))  # Same ID -> only the content has changed

1def modify_list(lst):
2    lst.append(100)
3
4my_list = [1, 2, 3]
5modify_list(my_list)
6print(my_list)  # [1, 2, 3, 100]

1def modify_number(n):
2    n += 10
3
4my_number = 5
5modify_number(my_number)
6print(my_number)  # 5 -> unchanged

 1# Bad example
 2def add_item(item, container=[]):
 3    container.append(item)
 4    return container
 5
 6print(add_item(1))  # [1]
 7print(add_item(2))  # [1, 2] -> unintended behavior
 8
 9# Good example
10def add_item(item, container=None):
11    if container is None:
12        container = []
13    container.append(item)
14    return container
15
16print(add_item(1))  # [1]
17print(add_item(2))  # [2]