对于下列Dog类,哪个叙述是错误的? class Dog { Dog(int m){ } Dog(double m){ } int Dog(int m){ return 23; } void Dog(double m){ } }A.Dog(int m)与Dog(double m)互为重载的构造方法。B.int Dog(int m)与void Dog(double m)互为重载的非构造方法。C.Dog类只有两个构造方法,而且没有无参数的构造方法。D.Dog类有3个构造方法。

对于下列Dog类,哪个叙述是错误的? class Dog { Dog(int m){ } Dog(double m){ } int Dog(int m){ return 23; } void Dog(double m){ } }

A.Dog(int m)与Dog(double m)互为重载的构造方法。

B.int Dog(int m)与void Dog(double m)互为重载的非构造方法。

C.Dog类只有两个构造方法,而且没有无参数的构造方法。

D.Dog类有3个构造方法。


参考答案和解析
Dog类有3个构造方法。

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