在Python中修改闭包的绑定变量 [英] Modify bound variables of a closure in Python

问题描述

有没有办法修改闭包中的一个变量的绑定值？

  def foo（）：
 var_a = 2 
 var_b = 3 
 
 def _closure（x）：
 return var_a + var_b + x 
 
 return _closure 
 
 $ b localClosure = foo（）
 
＃本地闭包现在是return 2 + 3 + x
a = localClosure（1）＃2 + 3 + 1 == 6 
 
 ＃DO某些魔法在这里把var_a的闭包变成0 
＃...但是什么魔术？这是可能吗？ 
 
＃本地闭包现在是return 0 + 3 + x
b = localClosure（1）＃0 + 3 +1 == 4 
  
 
 
解决方案
我认为在Python中没有办法。当定义闭包时，捕获包围作用域中的变量的当前状态，并且不再具有可直接引用的名称（从闭包外部）。如果你再次调用 foo（），新的闭包将有一个不同的变量从包围范围。
 
 
 
在你的简单例子中，你最好使用类：
  class foo：
 def __init__ （self）：
 self.var_a = 2 
 self.var_b = 3 
 
 def __call __（self，x）：
 return self.var_a + self.var_b + x 
 
 localClosure = foo（）
 
＃本地闭包现在是return 2 + 3 + x
a = localClosure（1）＃2 + 3 + 1 == 6 
 
＃让一些魔法把闭包的var_a变成0 
＃...但是什么魔术？这是可能吗？ 
 localClosure.var_a = 0 
 
＃本地闭包现在为return 0 + 3 + x
b = localClosure（1）＃0 + 3 +1 == 4 
  
如果你使用这种技术，我将不再使用 localClosure ，因为它不再是一个闭包。但是，它的工作原理与一样。
 
Is there any way to modify the bound value of one of the variables inside a closure? Look at the example to understand it better.
def foo():
    var_a = 2
    var_b = 3

    def _closure(x):
        return var_a + var_b + x

    return _closure


localClosure = foo()

# Local closure is now "return 2 + 3 + x"
a = localClosure(1) # 2 + 3 + 1 == 6

# DO SOME MAGIC HERE TO TURN "var_a" of the closure into 0
# ...but what magic? Is this even possible?

# Local closure is now "return 0 + 3 + x"
b = localClosure(1) # 0 + 3 +1 == 4

解决方案
I don't think there is any way to do that in Python. When the closure is defined, the current state of variables in the enclosing scope is captured and no longer has a directly referenceable name (from outside the closure). If you were to call foo() again, the new closure would have a different set of variables from the enclosing scope.

In your simple example, you might be better off using a class:
class foo:
        def __init__(self):
                self.var_a = 2
                self.var_b = 3

        def __call__(self, x):
                return self.var_a + self.var_b + x

localClosure = foo()

# Local closure is now "return 2 + 3 + x"
a = localClosure(1) # 2 + 3 + 1 == 6

# DO SOME MAGIC HERE TO TURN "var_a" of the closure into 0
# ...but what magic? Is this even possible?
localClosure.var_a = 0

# Local closure is now "return 0 + 3 + x"
b = localClosure(1) # 0 + 3 +1 == 4
If you do use this technique I would no longer use the name localClosure because it is no longer actually a closure. However, it works the same as one.

                        
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