35 Python __call__

如果一个对象定义了__call__(slef, [,*args [,**kwargs]])，则该对象可以模拟函数的行为，像函数一样使用它(相当于重载了括号运算符…)。

即x(*args,**kwargs)等同于调用了x.__call__(*args,**kwargs)

所以如果想将一个类实例变成一个可调用对象，只需实现__call__()方法即可。

#!/usr/bin/python
# -*- coding: utf-8 -*-
class Foo(object):
    def __call__(self, x):
        print "__call__ x  = %s"%x

f = Foo()
f(1)            #__call__ x  = 1

class Next:
  List = []
  def __init__(self,low,high):
    for i in range(low,high):
      self.List.append(i)

  def __call__(self,x):
    if x < len(self.List):
        return self.List[x]
    return 'index out of range'

b = Next(1,7)   #[1, 4, 9, 16, 25, 36]
print b(2)      #3
print b(10)     #index out of range

problem，为什么a()和a.__call__ ()的结果不一样?

class A(object):
    def __call__(self):
        print("__call__ from A!")
a = A()
a()                    #__call__ from A
a.__call__ = lambda : "__call__ from lambda"
print a.__call__()     #__call__ from lambda 
a()                    #__call__ from A

原因在于,在Python中,新式类的内嵌特殊方法和实例的属性字典是相互隔离的…

官网解释

For new - style classes , implicit invocations of special methods are only guaranteed to work correctly if defined on an object ‘s type , not in the object ’ s instance dictionary. That behaviour is the reason why the following code raises an exception(unlike the equivalent example with old - style classes)

当执行a.__call__ = lambda :"__call__ from lambda "时，在a.__dict__中新增了key为'__call__'的成员，但是当执行a()时，因为涉及特殊方法的调用，不会从 a.__dict__中寻找属性，而是从type(a).__dict__中寻找属性。
因此,就会出现如上所述的情况。

验证

a = A()
print a.__dict__       #{}
a()                    #__call__ from A
a.__call__ = lambda : "__call__ from lambda"
print a.__dict__       #{'__call__':  at 0x0000000002F6E908>},相当于实例方法
print a.__call__()     #__call__ from lambda,从实例字典中查找的 
a()                    #__call__ from A,从类字典中查找的

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