Python Closure：为什么在更改数字时允许更改闭合字典？
当我尝试更改闭合
def a():
x = {'a':'b'}
def b():
x['a'] = 'd'
print(x)
return b
>>> b = a()
>>> b()
{'a':'d'}
输出中的字典时，该代码效果很好，可以满足我的期望。但是，为什么下面的代码不起作用？
def m():
x = 1
def n():
x += 1
print(x)
return n
>>> n = m()
>>> n()
unboundlocalerror：locultlocalerror：locallible'x'在分配之前引用了
老实说，我知道我们可以使用非局部x语句来解决此问题
但是有人可以对我更深入地解释原因吗？ 字典和数字之间的差异
谢谢！
this code works well when I try to change dictionary in closure
def a():
x = {'a':'b'}
def b():
x['a'] = 'd'
print(x)
return b
>>> b = a()
>>> b()
{'a':'d'}
output is meeting my expectation. but why the code below doesn't work?
def m():
x = 1
def n():
x += 1
print(x)
return n
>>> n = m()
>>> n()
UnboundLocalError: local variable 'x' referenced before assignment
Honestly, I've known that we can use nonlocal x statement to solve this problem
But can anybody explain the reason more deeply for me? what the difference between a dictionary and a number
Thanks!
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python有一个很好的 FAQ 专门针对此。
简而>对象，您修改相同的对象，因此您不会重新分配变量。如果整数是不可变的，则通过执行
+=创建一个新对象并将其放入x中。由于X现在已在内部函数内定义，但是您正在尝试从外部功能带来数据，因此您有问题。您可以使用
id（）检查是否是同一对象。Python has a great FAQ specifically on this.
In short, when you modify a dictionary, or any mutable object, you modify the same object, so you don't re-assign the variable. In case of an integer, since it's immutable, by doing a
+=you create a new object and put it intox. Since x is now defined inside the inner function, but you're trying to bring data from the outer function, you have an issue.You can check if it's the same object using
id().