Java泛型参数与泛型参数的基础

问题描述

<$ p

我想知道是否有一个优雅的解决方案可以在Java中执行此操作（除了显而易见的 - 声明不同的/显式函数外）。 $ b $ private static HashMap< String，Integer> nameStringIndexMap
= new HashMap< String，Integer>（）;
private static HashMap< Buffer，Integer> nameBufferIndexMap
= new HashMap< Buffer，Integer>（）;

//和函数
private static String newName（Object object，
HashMap< Object，Integer> nameIndexMap）{
....
}

问题是我无法传递 nameStringIndexMap 或 nameBufferIndexMap 参数。我没有关于更明智的解决方案，除了做另一个显式地需要 HashMap< String，Integer> 或 HashMap< Buffer，Integer> 参数。

我的问题是：
在一个更优雅的解决方案/使用泛型或类似的东西？

谢谢，

Iulian

解决方案

您可以使您的函数具有通用性：

  private static< E extends Object> String newName（E object，
 HashMap< E，Integer> nameIndexMap）{
 .... 
} 
  

这将函数的两个参数绑定在一起，因此对于 HashMap< String，Integer> ，您只能传递 String 实例作为第一个参数。这可能是也可能不是你想要的：如果你只想从地图获取元素，Jon的解决方案更简单，但是如果你想将这个对象添加到地图中，这是唯一的选择。

I am wondering if there's an elegant solution for doing this in Java (besides the obvious one - of declaring a different/explicit function. Here is the code:

private static HashMap<String, Integer> nameStringIndexMap
        = new HashMap<String, Integer>();
private static HashMap<Buffer, Integer> nameBufferIndexMap
        = new HashMap<Buffer, Integer>();

// and a function
private static String newName(Object object,
        HashMap<Object, Integer> nameIndexMap){
    ....
}

The problem is that I cannot pass nameStringIndexMap or nameBufferIndexMap parameters to the function. I don't have an idea about a more elegant solution beside doing another function which explicitly wants a HashMap<String, Integer> or HashMap<Buffer, Integer> parameter.

My question is:Can this be made in a more elegant solution/using generics or something similar?

Thank you,

Iulian

解决方案

You could make your function generic too:

private static <E extends Object> String newName(E object,
        HashMap<E, Integer> nameIndexMap){
    ....
}

This bounds the two parameters of the function together, so for a HashMap<String, Integer> you can only pass String instances as first parameter. This may or may not be what you exactly want: if you only want to get elements from the map, Jon's solution is simpler, but if you want to add this object to the map, this one is the only choice.

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