可以在Java数组作为一个HashMap的关键

问题描述

如果一个HashMap的键是一个字符串数组：

If a HashMap's key is a string array:

HashMap<String[], String> pathMap;

您可以通过使用一个新创建的字符串数组或是否必须是同一个String []对象？访问地图

Can you access the map by using a newly created string array or does it have to be the same String[] object?

pathMap = new HashMap<>(new String[] { "korey", "docs" }, "/home/korey/docs");
String path = pathMap.get(new String[] { "korey", "docs" });

推荐答案

它必须是相同的对象。 A 的HashMap 比较了使用键的equals（）和Java的两个数组仅当它们是同一个对象相等。

It will have to be the same object. A HashMap compares keys using equals() and two arrays in Java are equal only if they are the same object.

如果你想要的值相等，然后写你自己的容器类，包装了一个的String [] ，并提供了相应的语义等于（）和散code（）。在这种情况下，这将是最好的，使容器不可改变的，如改变哈希code为对象，基于散列的容器类严重破坏。

If you want value equality, then write your own container class that wraps a String[] and provides the appropriate semantics for equals() and hashCode(). In this case, it would be best to make the container immutable, as changing the hash code for an object plays havoc with the hash-based container classes.

修改

正如其他人所指出的那样，列表与LT;弦乐＆GT; 有你似乎想为容器对象的语义。所以，你可以做这样的事情：

As others have pointed out, List<String> has the semantics you seem to want for a container object. So you could do something like this:

HashMap<List<String>, String> pathMap;

pathMap.put(
    // unmodifiable so key cannot change hash code
    Collections.unmodifiableList(Arrays.asList("korey", "docs")),
    "/home/korey/docs"
);

// later:
String dir = pathMap.get(Arrays.asList("korey", "docs"));

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