是否保证新的Integer（i）== i在Java？

问题描述

请考虑以下代码段：

  int i = 99999999; 
 byte b = 99; 
 short s = 9999; 
整数ii = Integer.valueOf（9）; //应该在缓存内
 
 System.out.println（new Integer（i）== i）; //true
 System.out.println（new Integer（b）== b）; //true
 System.out.println（new Integer（s）== s）; //true
 System.out.println（new Integer（ii）== ii）; //false
  

很明显为什么最后一行会始终 false：我们使用 == 引用标识比较和 / code>对象将永远不为已经存在的对象 == 。

问题是前三行： 比较 int 与整数自动取消装箱？有没有情况下，原语将被自动装箱，而不是参考身份比较执行？ （这将全部 false ！）

解决方案

指定二进制数字促销的规则。部分：

二进制数字促销适用于多个数字运算符，包括数值相等运算符==和！=。

（数字等式运算符==和！=）指定：

相反，（参考均等运算符==和！=） p>

这符合对装箱和取消装箱的共同理解，只有当不匹配时才会这样做。 p>

Consider the following snippet:

    int i = 99999999;
    byte b = 99;
    short s = 9999;
    Integer ii = Integer.valueOf(9); // should be within cache

    System.out.println(new Integer(i) == i); // "true"
    System.out.println(new Integer(b) == b); // "true"
    System.out.println(new Integer(s) == s); // "true"
    System.out.println(new Integer(ii) == ii); // "false"

It's obvious why the last line will ALWAYS prints "false": we're using == reference identity comparison, and a new object will NEVER be == to an already existing object.

The question is about the first 3 lines: are those comparisons guaranteed to be on the primitive int, with the Integer auto-unboxed? Are there cases where the primitive would be auto-boxed instead, and reference identity comparisons are performed? (which would all then be false!)

解决方案

Yes. JLS §5.6.2 specifies the rules for binary numeric promotion. In part:

Binary numeric promotion applies for several numeric operators, including "the numerical equality operators == and !=."

JLS §15.21.1 (Numerical Equality Operators == and !=) specifies:

In contrast, JLS §15.21.3 (Reference Equality Operators == and !=) provides:

This fits the common understanding of boxing and unboxing, that's it only done when there's a mismatch.

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