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。
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
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<T, U> can still return <code>true</code> for method <code>equals</code>
* called with a Pair<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<T, U></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 <v1, v2> of
* this pair and <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<T, U> can return <code>true</code> for equality testing with
* a Pair<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?
其他回答
虽然这个问题已经有十多年的历史了,但我觉得有必要提一下,从Java 14开始,Records可以为这个问题提供非常简单和轻量级的解决方案,而不需要任何形式的外部库或依赖项。
例如,以下记录类声明将是实现所需功能所需的全部内容:
record Pair<K, V>(K key, V value) { }
这样一个记录类可以这样使用:
// Declare a pair object containing two integers
var integerIntegerPair = new Pair<>(1, 2);
// Declare a pair object containing a String and an integer
var stringIntegerPair = new Pair<>("String", 20);
// Declare a pair object containing two other pairs!
var pairPairPair = new Pair<>(new Pair<>(1, 2), new Pair<>("String", 20));
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
试试VAVR元组。
vavr不仅有一组很好的元组类型,而且对函数式编程也有很好的支持。
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。
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<T, U> can still return <code>true</code> for method <code>equals</code>
* called with a Pair<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<T, U></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 <v1, v2> of
* this pair and <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<T, U> can return <code>true</code> for equality testing with
* a Pair<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?