如何修补`__call__`方法？

Question

我似乎无法修补__call__类实例的方法(是的,我想修补单个实例,而不是所有实例).

以下代码:

class A(object):
    def test(self):
        return "TEST"

    def __call__(self):
        return "EXAMPLE"

a = A()
print("call method: {0}".format(a.__call__))
print("test method: {0}".format(a.test))
a.__call__ = lambda : "example"
a.test = lambda : "test"
print("call method: {0}".format(a.__call__))
print("test method: {0}".format(a.test))

print(a())
print("Explicit call: {0}".format(a.__call__()))
print(a.test())

输出:

call method: <bound method A.__call__ of <__main__.A object at 0x7f3f2d60b6a0>>
test method: <bound method A.test of <__main__.A object at 0x7f3f2d60b6a0>>
call method: <function <lambda> at 0x7f3f2ef4ef28>
test method: <function <lambda> at 0x7f3f2d5f8f28>
EXAMPLE
Explicit call: example
test

虽然我想输出:

...
example
Explicit call: example
test

我如何monkeypatch __call__()？为什么我不能像修补其他方法那样修补它？

虽然这个答案告诉我们该怎么做(据说,我还没有测试过),但它没有解释为什么这个问题的部分原因.

Answer 1

所以,正如JJ Hakala所评论的那样,Python真正做的是调用:

type(a).__call__(a)

因此,如果我想覆盖该__call__方法,我必须覆盖__call__一个类,但如果我不想影响同一个类的其他实例的行为,我需要使用overriden __call__方法创建一个新类.

所以如何覆盖的示例__call__将如下所示:

class A(object):
    def test(self):
        return "TEST"

    def __call__(self):
        return "EXAMPLE"

def patch_call(instance, func):
    class _(type(instance)):
        def __call__(self, *arg, **kwarg):
           return func(*arg, **kwarg)
    instance.__class__ = _

a = A()
print("call method: {0}".format(a.__call__))
print("test method: {0}".format(a.test))
patch_call(a, lambda : "example")
a.test = lambda : "test"
print("call method: {0}".format(a.__call__))
print("test method: {0}".format(a.test))

print("{0}".format(a()))
print("Explicit a.__call__: {0}".format(a.__call__()))
print("{0}".format(a.test()))

print("Check instance of a: {0}".format(isinstance(a, A)))

运行它会产生以下输出:

call method: <bound method A.__call__ of <__main__.A object at 0x7f404217a5f8>>
test method: <bound method A.test of <__main__.A object at 0x7f404217a5f8>>
call method: <bound method patch_call.<locals>._.__call__ of <__main__.patch_call.<locals>._ object at 0x7f404217a5f8>>
test method: <function <lambda> at 0x7f404216d048>
example
Explicit a.__call__: example
test
Check instance of a: True 

Answer 2

对于自定义类，只有在对对象的类型（而不是在对象的实例字典中）定义的情况下，才能保证对特殊方法的隐式调用可以正常工作。该行为是以下代码引发异常的原因：

>>> class C:
...     pass
...
>>> c = C()
>>> c.__len__ = lambda: 5
>>> len(c)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: object of type 'C' has no len()

来源：https : //docs.python.org/3/reference/datamodel.html#special-lookup