Java中没有Pair<L,R>,有什么好理由吗?与这个c++构造等价的是什么?我宁愿避免重新实现我自己的。

似乎1.6提供了类似的东西(AbstractMap.SimpleEntry<K,V>),但这看起来相当复杂。


当前回答

您可以使用javafx实用程序类Pair,它的作用与c++中的Pair <>相同。https://docs.oracle.com/javafx/2/api/javafx/util/Pair.html

其他回答

Collections.singletonMap(left, rigth);
public class Pair<K, V> {

    private final K element0;
    private final V element1;

    public static <K, V> Pair<K, V> createPair(K key, V value) {
        return new Pair<K, V>(key, value);
    }

    public Pair(K element0, V element1) {
        this.element0 = element0;
        this.element1 = element1;
    }

    public K getElement0() {
        return element0;
    }

    public V getElement1() {
        return element1;
    }

}

用法:

Pair<Integer, String> pair = Pair.createPair(1, "test");
pair.getElement0();
pair.getElement1();

不可变,只有一对!

实现Pair with的另一种方法。

Public immutable fields, i.e. simple data structure. Comparable. Simple hash and equals. Simple factory so you don't have to provide the types. e.g. Pair.of("hello", 1); public class Pair<FIRST, SECOND> implements Comparable<Pair<FIRST, SECOND>> { public final FIRST first; public final SECOND second; private Pair(FIRST first, SECOND second) { this.first = first; this.second = second; } public static <FIRST, SECOND> Pair<FIRST, SECOND> of(FIRST first, SECOND second) { return new Pair<FIRST, SECOND>(first, second); } @Override public int compareTo(Pair<FIRST, SECOND> o) { int cmp = compare(first, o.first); return cmp == 0 ? compare(second, o.second) : cmp; } // todo move this to a helper class. private static int compare(Object o1, Object o2) { return o1 == null ? o2 == null ? 0 : -1 : o2 == null ? +1 : ((Comparable) o1).compareTo(o2); } @Override public int hashCode() { return 31 * hashcode(first) + hashcode(second); } // todo move this to a helper class. private static int hashcode(Object o) { return o == null ? 0 : o.hashCode(); } @Override public boolean equals(Object obj) { if (!(obj instanceof Pair)) return false; if (this == obj) return true; return equal(first, ((Pair) obj).first) && equal(second, ((Pair) obj).second); } // todo move this to a helper class. private boolean equal(Object o1, Object o2) { return o1 == null ? o2 == null : (o1 == o2 || o1.equals(o2)); } @Override public String toString() { return "(" + first + ", " + second + ')'; } }

根据Java语言的性质,我认为人们实际上并不需要Pair,通常他们需要的是一个接口。这里有一个例子:

interface Pair<L, R> {
    public L getL();
    public R getR();
}

所以,当人们想要返回两个值时,他们可以这样做:

... //Calcuate the return value
final Integer v1 = result1;
final String v2 = result2;
return new Pair<Integer, String>(){
    Integer getL(){ return v1; }
    String getR(){ return v2; }
}

This is a pretty lightweight solution, and it answers the question "What is the semantic of a Pair<L,R>?". The answer is, this is an interface build with two (may be different) types, and it has methods to return each of them. It is up to you to add further semantic to it. For example, if you are using Position and REALLY want to indicate it in you code, you can define PositionX and PositionY that contains Integer, to make up a Pair<PositionX,PositionY>. If JSR 308 is available, you may also use Pair<@PositionX Integer, @PositionY Ingeger> to simplify that.

编辑: 这里我应该指出的一点是,上面的定义显式地将类型参数名和方法名联系起来。这是对那些认为Pair缺乏语义信息的人的回答。实际上,getL方法的意思是“给我对应于类型参数L的类型的元素”,这确实意味着什么。

编辑: 下面是一个简单的实用程序类,可以让生活变得更简单:

class Pairs {
    static <L,R> Pair<L,R> makePair(final L l, final R r){
        return new Pair<L,R>(){
            public L getL() { return l; }
            public R getR() { return r; }   
        };
    }
}

用法:

return Pairs.makePair(new Integer(100), "123");

Pair是一个很好的东西,作为一个复杂泛型的基本构造单元,例如,这是我的代码:

WeakHashMap<Pair<String, String>, String> map = ...

它和Haskell的元组是一样的