爲什麼Python中有不可變對象？

Question

所以Python是通過引用。總是。但是像整數，字符串和元組這樣的對象，即使傳入一個函數，也不能改變（因此它們被稱爲不可變的）。這是一個例子。

def foo(i, l): 
    i = 5 # this creates a new variable which is also called i 
    l = [1, 2] # this changes the existing variable called l 

i = 10 
l = [1, 2, 3] 
print(i) 
# i = 10 
print(l) 
# l = [1, 2, 3] 
foo(i, l) 
print(i) 
# i = 10 # variable i defined outside of function foo didn't change 
print(l) 
# l = [1, 2] # l is defined outside of function foo did change 

所以你可以看到整數對象是不可變的，而列表對象是可變的。

甚至在Python中有不可變對象的原因是什麼？如果所有對象都是可變的，那麼像Python這樣的語言會有什麼優點和缺點？

Answer 1

你的例子不正確。 l沒有更改foo()的範圍以外。 i和l裏面的foo()是指向新對象的新名稱。現在

Python 2.7.10 (default, Aug 22 2015, 20:33:39) 
[GCC 4.2.1 Compatible Apple LLVM 7.0.0 (clang-700.0.59.1)] on darwin 
Type "help", "copyright", "credits" or "license" for more information. 
>>> def foo(i, l): 
... i = 5 # this creates a local name i that points to 5 
... l = [1, 2] # this creates a local name l that points to [1, 2] 
... 
>>> i = 10 
>>> l = [1, 2, 3] 
>>> print(i) 
10 
>>> print(l) 
[1, 2, 3] 
>>> foo(i, l) 
>>> print(i) 
10 
>>> print(l) 
[1, 2, 3] 

，如果你改變foo()變異l，這是一個不同的故事

>>> def foo(i, l): 
...  l.append(10) 
... 
>>> foo(i, l) 
>>> print(l) 
[1, 2, 3, 10] 

Python 3的實例相同

Python 3.4.3 (v3.4.3:9b73f1c3e601, Feb 23 2015, 02:52:03) 
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin 
Type "help", "copyright", "credits" or "license" for more information. 
>>> def foo(i, l): 
... i = 5 # this creates a new variable which is also called i 
... l = [1, 2] # this changes the existing variable called l 
... 
>>> i = 10 
>>> l = [1, 2, 3] 
>>> print(i) 
10 
>>> print(l) 
[1, 2, 3] 
>>> foo(i, l) 
>>> print(i) 
10 
>>> print(l) 
[1, 2, 3]