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saximi
V2EX  ›  Python

请问类名()何时会等于类名.__call__() ?

  •  
  •   saximi · Aug 29, 2017 · 1986 views
    This topic created in 3191 days ago, the information mentioned may be changed or developed.
    
    class Foo: 
            def __call__(): 
                    print("call") 
    
    if __name__=='__main__': 
            Foo() 
            print("--------") 
            Foo.__call__() 
    
    
    上面代码输出如下:
    --------
    call
    
    

    请问为何这里 Foo()不等于 Foo.call(),什么情况下才会相等呢? 谢谢

    4 replies    2017-08-30 21:11:20 +08:00
    CodeDrift
        1
    CodeDrift  
       Aug 29, 2017
    Foo() 执行的是__init__方法 可以试试这样 f = Foo() f()
    janxin
        3
    janxin  
       Aug 29, 2017 via iPad
    class Foo:
    def __call__(self):
    print("call")

    if __name__=='__main__':
    f = Foo()
    print("--------")
    f()
    saximi
        4
    saximi  
    OP
       Aug 30, 2017
    @janxin 谢谢,我还是有疑问,为何__call__加入 self 参数后就使得 Foo()会调用 Foo.__call__() ,否则这不会调用呢?
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