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>EXAMPLEExplicit call: exampletest
虽然我想输出:
...exampleExplicit call: exampletest
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this answer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.
解决方法 所以,正如 J.J. Hakala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__call__方法,我必须覆盖类的__call__,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overrIDe__call__创建一个新类.方法.
所以如何覆盖__call__的示例如下所示:
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>exampleExplicit a.__call__: exampletestCheck instance of a: True总结
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