python - 与语言 X 闭包相比，Python 中的闭包有哪些限制？

Question

其中 X 是任何支持某种闭包风格的编程语言(C#、Javascript、Lisp、Perl、Ruby、Scheme 等)。

Closures in Python 中提到了一些限制。 (与 Ruby 的闭包相比)，但这篇文章已经过时，现代 Python 中不再存在许多限制。

查看具体限制的代码示例会很棒。

相关问题:

• Can you explain closures (as they relate to Python)?
• What is a ‘Closure’?
• How does a javascript closure work ?

Answer 1

目前最重要的限制是您不能分配给外部范围变量。换句话说，闭包是只读的:

>>> def outer(x): 
...     def inner_reads():
...         # Will return outer's 'x'.
...         return x
...     def inner_writes(y):
...         # Will assign to a local 'x', not the outer 'x'
...         x = y
...     def inner_error(y):
...         # Will produce an error: 'x' is local because of the assignment,
...         # but we use it before it is assigned to.
...         tmp = x
...         x = y
...         return tmp
...     return inner_reads, inner_writes, inner_error
... 
>>> inner_reads, inner_writes, inner_error = outer(5)
>>> inner_reads()
5
>>> inner_writes(10)
>>> inner_reads()
5
>>> inner_error(10)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 11, in inner_error
UnboundLocalError: local variable 'x' referenced before assignment

除非另有声明，否则在本地范围(函数)中分配的名称始终是本地的。虽然有一个“全局”声明来声明一个全局变量，即使它被分配给它，但对于封闭的变量，还没有这样的声明。在 Python 3.0 中，有(将有)'nonlocal' 声明可以做到这一点。

您可以同时使用可变容器类型来解决此限制:

>>> def outer(x):
...     x = [x]
...     def inner_reads():
...         # Will return outer's x's first (and only) element.
...         return x[0]
...     def inner_writes(y):
...         # Will look up outer's x, then mutate it.      
...         x[0] = y
...     def inner_error(y):
...         # Will now work, because 'x' is not assigned to, just referenced.
...         tmp = x[0]
...         x[0] = y
...         return tmp
...     return inner_reads, inner_writes, inner_error
... 
>>> inner_reads, inner_writes, inner_error = outer(5)
>>> inner_reads()
5
>>> inner_writes(10)
>>> inner_reads()
10
>>> inner_error(15)
10
>>> inner_reads()
15