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

为了方便起见,以下是一些具有多级元组的库:

JavaTuples。它只有1-10度的元组。 JavaSlang。0-8度的元组和许多其他功能性的东西。 jOOλ。0-16度的元组和其他一些功能性的好东西。(免责声明,我在维修公司工作) Functional Java。0-8度的元组和许多其他功能性的东西。

已经提到的其他库至少包含Pair元组。

具体来说,在函数式编程的上下文中,它使用了大量的结构类型,而不是名义类型(正如公认答案中提倡的那样),这些库及其元组非常方便。

其他回答

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");

我注意到所有的Pair实现都散布在这里,属性含义取决于两个值的顺序。当我想到一对时,我想到的是两件物品的组合,这两件物品的顺序不重要。下面是我对一个无序对的实现,使用hashCode和equals重写以确保集合中的期望行为。也可克隆。

/**
 * The class <code>Pair</code> models a container for two objects wherein the
 * object order is of no consequence for equality and hashing. An example of
 * using Pair would be as the return type for a method that needs to return two
 * related objects. Another good use is as entries in a Set or keys in a Map
 * when only the unordered combination of two objects is of interest.<p>
 * The term "object" as being a one of a Pair can be loosely interpreted. A
 * Pair may have one or two <code>null</code> entries as values. Both values
 * may also be the same object.<p>
 * Mind that the order of the type parameters T and U is of no importance. A
 * Pair&lt;T, U> can still return <code>true</code> for method <code>equals</code>
 * called with a Pair&lt;U, T> argument.<p>
 * Instances of this class are immutable, but the provided values might not be.
 * This means the consistency of equality checks and the hash code is only as
 * strong as that of the value types.<p>
 */
public class Pair<T, U> implements Cloneable {

    /**
     * One of the two values, for the declared type T.
     */
    private final T object1;
    /**
     * One of the two values, for the declared type U.
     */
    private final U object2;
    private final boolean object1Null;
    private final boolean object2Null;
    private final boolean dualNull;

    /**
     * Constructs a new <code>Pair&lt;T, U&gt;</code> with T object1 and U object2 as
     * its values. The order of the arguments is of no consequence. One or both of
     * the values may be <code>null</code> and both values may be the same object.
     *
     * @param object1 T to serve as one value.
     * @param object2 U to serve as the other value.
     */
    public Pair(T object1, U object2) {

        this.object1 = object1;
        this.object2 = object2;
        object1Null = object1 == null;
        object2Null = object2 == null;
        dualNull = object1Null && object2Null;

    }

    /**
     * Gets the value of this Pair provided as the first argument in the constructor.
     *
     * @return a value of this Pair.
     */
    public T getObject1() {

        return object1;

    }

    /**
     * Gets the value of this Pair provided as the second argument in the constructor.
     *
     * @return a value of this Pair.
     */
    public U getObject2() {

        return object2;

    }

    /**
     * Returns a shallow copy of this Pair. The returned Pair is a new instance
     * created with the same values as this Pair. The values themselves are not
     * cloned.
     *
     * @return a clone of this Pair.
     */
    @Override
    public Pair<T, U> clone() {

        return new Pair<T, U>(object1, object2);

    }

    /**
     * Indicates whether some other object is "equal" to this one.
     * This Pair is considered equal to the object if and only if
     * <ul>
     * <li>the Object argument is not null,
     * <li>the Object argument has a runtime type Pair or a subclass,
     * </ul>
     * AND
     * <ul>
     * <li>the Object argument refers to this pair
     * <li>OR this pair's values are both null and the other pair's values are both null
     * <li>OR this pair has one null value and the other pair has one null value and
     * the remaining non-null values of both pairs are equal
     * <li>OR both pairs have no null values and have value tuples &lt;v1, v2> of
     * this pair and &lt;o1, o2> of the other pair so that at least one of the
     * following statements is true:
     * <ul>
     * <li>v1 equals o1 and v2 equals o2
     * <li>v1 equals o2 and v2 equals o1
     * </ul>
     * </ul>
     * In any other case (such as when this pair has two null parts but the other
     * only one) this method returns false.<p>
     * The type parameters that were used for the other pair are of no importance.
     * A Pair&lt;T, U> can return <code>true</code> for equality testing with
     * a Pair&lt;T, V> even if V is neither a super- nor subtype of U, should
     * the the value equality checks be positive or the U and V type values
     * are both <code>null</code>. Type erasure for parameter types at compile
     * time means that type checks are delegated to calls of the <code>equals</code>
     * methods on the values themselves.
     *
     * @param obj the reference object with which to compare.
     * @return true if the object is a Pair equal to this one.
     */
    @Override
    public boolean equals(Object obj) {

        if(obj == null)
            return false;

        if(this == obj)
            return true;

        if(!(obj instanceof Pair<?, ?>))
            return false;

        final Pair<?, ?> otherPair = (Pair<?, ?>)obj;

        if(dualNull)
            return otherPair.dualNull;

        //After this we're sure at least one part in this is not null

        if(otherPair.dualNull)
            return false;

        //After this we're sure at least one part in obj is not null

        if(object1Null) {
            if(otherPair.object1Null) //Yes: this and other both have non-null part2
                return object2.equals(otherPair.object2);
            else if(otherPair.object2Null) //Yes: this has non-null part2, other has non-null part1
                return object2.equals(otherPair.object1);
            else //Remaining case: other has no non-null parts
                return false;
        } else if(object2Null) {
            if(otherPair.object2Null) //Yes: this and other both have non-null part1
                return object1.equals(otherPair.object1);
            else if(otherPair.object1Null) //Yes: this has non-null part1, other has non-null part2
                return object1.equals(otherPair.object2);
            else //Remaining case: other has no non-null parts
                return false;
        } else {
            //Transitive and symmetric requirements of equals will make sure
            //checking the following cases are sufficient
            if(object1.equals(otherPair.object1))
                return object2.equals(otherPair.object2);
            else if(object1.equals(otherPair.object2))
                return object2.equals(otherPair.object1);
            else
                return false;
        }

    }

    /**
     * Returns a hash code value for the pair. This is calculated as the sum
     * of the hash codes for the two values, wherein a value that is <code>null</code>
     * contributes 0 to the sum. This implementation adheres to the contract for
     * <code>hashCode()</code> as specified for <code>Object()</code>. The returned
     * value hash code consistently remain the same for multiple invocations
     * during an execution of a Java application, unless at least one of the pair
     * values has its hash code changed. That would imply information used for 
     * equals in the changed value(s) has also changed, which would carry that
     * change onto this class' <code>equals</code> implementation.
     *
     * @return a hash code for this Pair.
     */
    @Override
    public int hashCode() {

        int hashCode = object1Null ? 0 : object1.hashCode();
        hashCode += (object2Null ? 0 : object2.hashCode());
        return hashCode;

    }

}

这个实现已经经过了适当的单元测试,并且在Set和Map中的使用已经经过了尝试。

请注意,我并没有要求在公共领域发布这个。这是我为在应用程序中使用而编写的代码,因此如果您打算使用它,请避免直接复制,并在注释和名称上搞得一团糟。明白我的意思吗?

在Lombok的新版本中,你可以编译这个可爱的类:

@Value(staticConstructor = "of") public class Pair <E> {
  E first, second;
}

并像这样使用它:Pair<Value> pairOfValues = Pair。(value1, value2);

正如许多人已经指出的那样,Pair类是否有用实际上取决于用例。

我认为,对于私有帮助函数,如果使用Pair类可以使代码更具可读性,并且不值得花费精力创建另一个包含所有锅炉代码的值类,那么使用Pair类是完全合法的。

另一方面,如果您的抽象级别要求您清楚地记录包含两个对象或值的类的语义,那么您应该为它编写一个类。如果数据是业务对象,通常就是这种情况。

一如既往,这需要熟练的判断。

对于你的第二个问题,我推荐Apache Commons库中的Pair类。这些可能被认为是Java的扩展标准库:

https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/tuple/Pair.html

你可能还想看看Apache Commons的EqualsBuilder、HashCodeBuilder和ToStringBuilder,它们简化了为业务对象编写值类的过程。

尽管语法相似,但Java和c++有非常不同的范例。像Java一样写c++是糟糕的c++,像c++一样写Java是糟糕的Java。

使用像Eclipse这样基于反射的IDE,编写“pair”类的必要功能是快速而简单的。创建类,定义两个字段,使用各种“Generate XX”菜单选项在几秒钟内填写类。如果你想要Comparable界面,也许你必须很快地输入一个“compareTo”。

由于c++语言中有单独的声明/定义选项,代码生成器并不是很好,所以手工编写小的实用程序类更加费时乏味。由于pair是一个模板,您不必为不使用的函数付费,并且typedef功能允许为代码分配有意义的类型名,因此关于“无语义”的反对意见并不成立。