Java：在Set中添加/删除short和integer元素时的不同输出 [英] Java: Different outputs when add/remove short and integer elements in a Set

问题描述

我不确定如何提出这个问题。但是，这两行代码有什么不同？

I wasn't sure on how to ask this question. But, what is the different between these 2 lines of code?

Set<Integer> a = new HashSet<Integer>();
for (int i = 0; i < 100; i++) {
    a.add(i);
    a.remove(i - 1);
}

System.out.println(a.size());

我预计 99 是输出

输出 1

Set<Short> a = new HashSet<Short>();
for (Short i = 0; i < 100; i++) {
    a.add(i);
    a.remove(i - 1);
}

System.out.println(a.size());

我预计 99 是输出

输出 100

推荐答案

表达式的类型 i - 1 是 int 因为整数算术表达式中的所有操作数都被加宽到至少 int 。 设置<短> 有添加（短）和删除（对象）所以删除调用不需要进行转换/自动装箱。因此，您试图从一组短中删除​​ Integer 。

The type of the expression i - 1 is int because all operands in an integer arithmetic expression are widened to at least int. Set<Short> has add(Short) and remove(Object) so there's no casting/autoboxing needed on the remove call. Therefore you are trying to remove Integers from a set of Shorts.

请注意，由于这个原因，宣布 Set< Number> 几乎没有意义：

Note that for this reason it almost never makes sense to declare a Set<Number>:

final Set<Number> ns = new HashSet<>();
final short s = 1;
ns.add(s);
ns.add(s+0);
ns.add(s+0L);
System.out.println(ns); // prints [1, 1, 1]

作为奖励回合，如果更改设置实现到 TreeSet ，魔法消失，它抛出一个 ClassCastException ，放弃了这个技巧。

As a bonus round, if you change the set implementation to TreeSet, the magic disappears and it throws a ClassCastException, giving away the trick.

在内心深处，这个问题与等式是对称关系这一事实有关，这种关系不能区分右手边和左手边。使用Java的单调度方法无法实现这些语义。

Deep down, this issue has to do with the fact that equality is a symmetric relation, which must not distinguish the right hand side from the left hand side. These semantics are impossible to achieve with Java's single-dispatch methods.

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