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问题描述

我将值放入hashmap中,形式为

I'm putting values into the hashmap which is of the form,

Map<Long, Double> highLowValueMap=new HashMap<Long, Double>();
highLowValueMap.put(1l, 10.0);
highLowValueMap.put(2l, 20.0);

我想使用 values()创建一个列表地图方法。

List<Double> valuesToMatch=new ArrayList<>();
valuesToMatch=(List<Double>) highLowValueMap.values();

List<Double> valuesToMatch=(List<Double>) highLowValueMap.values();

但是,它会引发异常:

线程main中的异常java.lang.ClassCastException:

java.util.HashMap $值不能转换为java.util.List

但它允许我将它传递给一个列表的创建:

But it allows me to pass it in to the creation of a list:

List<Double> valuesToMatch  = new ArrayList<Double>( highLowValueMap.values());


推荐答案

TL; DR



TL;DR

List<V> al = new ArrayList<V>(hashMapVar.values());



说明



c> HashMap#values()返回 java.util.Collection< V> ,并且不能转换集合到 ArrayList ,因此你得到 ClassCastException 。

Explanation

Because HashMap#values() returns a java.util.Collection<V> and you can't cast a Collection into an ArrayList, thus you get ClassCastException.

我建议使用 ArrayList(Collection< ;? extends V>)构造函数。这个构造函数接受一个实现 Collection< ;?扩展V> 作为参数。当你传递 HashMap.values()的结果时,不会得到 ClassCastException :

I'd suggest using ArrayList(Collection<? extends V>) constructor. This constructor accepts an object which implements Collection<? extends V> as an argument. You won't get ClassCastException when you pass the result of HashMap.values() like this:

List<V> al = new ArrayList<V>(hashMapVar.values());



深入Java API源代码



HashMap#values():检查源中的返回类型,并问自己,是否可以将 java.util.Collection java.util.ArrayList ?否

Going further into the Java API source code

HashMap#values(): Check the return type in the source, and ask yourself, can a java.util.Collection be casted into java.util.ArrayList? No

public Collection<V> values() {
    Collection<V> vs = values;
    return (vs != null ? vs : (values = new Values()));
}

ArrayList(Collection):在源。参数是超类型的方法可以接受子类型吗?是

ArrayList(Collection): Check the argument type in the source. Can a method which argument is a super type accepts sub type? Yes

public ArrayList(Collection<? extends E> c) {
    elementData = c.toArray();
    size = elementData.length;
    // c.toArray might (incorrectly) not return Object[] (see 6260652)
    if (elementData.getClass() != Object[].class)
        elementData = Arrays.copyOf(elementData, size, Object[].class);
}

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10-20 09:26