我如何实例化？包含代码 [英] How I instantiate? Include code

问题描述

编译器不允许我在最后一行保留< X，Y>，我不明白为什么。

The compiler won't permit me to keep <X,Y> on the last line and I do not understand why.

我如何得到这样的要编译的通用构造？

How do I get such a generic construction to compile?

我尝试将代码更改为：

 X a = new A<X,Y>(); // "Type mismatch: cannot convert from A<X,Y> to X" 
 Y b = new B<X,Y>(); // "Type mismatch: cannot convert from B<X,Y> to Y" 

 W<X,Y> s = new M<X,Y>(a,b); // no error

我有点迷失 - 请帮忙！

I am a bit lost - please help!

推荐答案

的构造函数M< X，Y> 预计会收到 X 和 Y ，但你'试图给它一个 IA< X，Y> 和 IB< X，Y> 。必要的关系是相反的; X 是 IA< X，Y> ，但反之亦然， Y 类似。

The constructor of M< X, Y > expects to receive an X and a Y, but you're trying to give it an IA< X, Y > and an IB< X, Y >. The necessary relationships are reversed; X is an IA< X, Y >, but not vice-versa, and similarly for Y.

以下编译，但似乎对你所追求的内容没有足够的限制：

The following compiles, but appears to be not restrictive enough for what you are after:

class A<X extends IA<X,Y>, Y extends IB<X,Y>> implements IA<X,Y>{}
class B<X extends IA<X,Y>, Y extends IB<X,Y>> implements IB<X,Y>{}
interface IA<X extends IA<X,Y>, Y extends IB<X,Y>> {}
interface IB<X extends IA<X,Y>, Y extends IB<X,Y>> {}

class M<X extends IA<X,Y>, Y extends IB<X,Y>> extends W<X,Y>{
    public M(IA<X,Y> x, IB<X,Y> y){} // this is the only change
}

class W<X extends IA<X,Y>, Y extends IB<X,Y>> {}


//To my check class code:

public <X extends IA<X,Y>, Y extends IB<X,Y>> void check() {
    IA<X,Y> a = new A<X,Y>();
    IB<X,Y> b = new B<X,Y>();

    W<X,Y> s = new M<X,Y>(a,b);
}

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