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下列Dog类,错误的描述是: class Dog{ Dog(int m){} Dog(double m){} int Dog(int m){ return 1;} void Dog(double m){} }

A.Dog类有3个构造方法

B.Dog(int m)与 Dog(double m)互为重载的构造方法。

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

D.Dog类只有两个构造方法,而且没有无参数的构造方法。


参考答案和解析
Dog(int m)与 int Dog(int m) 是重载的构造方法;
更多 “下列Dog类,错误的描述是: class Dog{ Dog(int m){} Dog(double m){} int Dog(int m){ return 1;} void Dog(double m){} }A.Dog类有3个构造方法B.Dog(int m)与 Dog(double m)互为重载的构造方法。C.int Dog(int m)与 void Dog(double m)互为重载的非构造方法。D.Dog类只有两个构造方法,而且没有无参数的构造方法。” 相关考题
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