如何比较两个整数？ [英] How do I compare two Integers?

问题描述

我必须比较两个 Integer 对象（不是 int ）。

I have to compare two Integer objects (not int). What is the canonical way to compare them?

Integer x = ...
Integer y = ...

我可以想到这个：

if (x == y) 

== 运算符只比较引用，因此这将只适用于较小的整数值。

The == operator only compares references, so this will only work for lower integer values. But perhaps auto-boxing kicks in...?

if (x.equals(y)) 

这看起来像一个昂贵的操作。

This looks like an expensive operation. Are there any hash codes calculated this way?

if (x.intValue() == y.intValue())

有点冗长...

strong> EDIT：感谢您的回复。虽然我知道现在做什么，事实是分布在所有现有的答案（甚至删除的:)），我不知道，哪一个接受。所以我会接受最好的答案，它指所有三种比较的可能性，或至少前两个。

Thank you for your responses. Although I know what to do now, the facts are distributed on all of the existing answers (even the deleted ones :)) and I don't really know, which one to accept. So I'll accept the best answer, which refers to all three comparison possibilities, or at least the first two.

推荐答案

是equals方法所做的：

This is what the equals method does:

public boolean equals(Object obj) {
    if (obj instanceof Integer) {
        return value == ((Integer)obj).intValue();
    }
    return false;
}

正如你所看到的，没有哈希码计算，其他操作发生在那里。虽然 x.intValue（）== y.intValue（）可能会稍微快些，你进入微优化领域。加上编译器可能优化 equals（）调用，虽然我不知道这一定。

As you can see, there's no hash code calculation, but there are a few other operations taking place there. Although x.intValue() == y.intValue() might be slightly faster, you're getting into micro-optimization territory there. Plus the compiler might optimize the equals() call anyway, though I don't know that for certain.

我一般会使用原始 int ，但如果我不得不使用 Integer ，我会坚持使用 equals（）。

I generally would use the primitive int, but if I had to use Integer, I would stick with equals().

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