java - 字符串池行为

Question

我读了这个Questions about Java's String pool 并了解字符串池的基本概念，但仍然不了解其行为。

首先:如果你直接赋值并且 s1 和 s2 都引用池中的同一个对象，它就有效

String s1 = "a" + "bc";
String s2 = "ab" + "c";
System.out.println("s1 == s2? " + (s1 == s2));

但是如果我更改字符串 s1+="d"，那么池中应该有一个字符串对象“abcd”？那么当我更改 s2+="d"时，它应该在池中找到字符串对象“abcd”并将该对象分配给 s2？但事实并非如此，它们也不是指同一个对象。这是为什么？

String s1 = "abc";
String s2 = "abc";
System.out.println("s1 == s2? " + (s1 == s2));

s1 += "d";                  
s2 += "d";
System.out.println("s1 == s2? " + (s1 == s2));

Answer 1

当您在字符串上调用 String.intern() 时，字符串保证被合并。

String s1 = "abcd".intern();
String s2 = "abc";
s2 += "d";
s2 = s2.intern();
s1 == s2 // returns true

当编译器看到一个常量时，它会足够聪明地优化和合并字符串文字，即:

String s1 = "abcd";
String s2 = "abcd";
s1 == s2 // returns true

Java Language Specification状态:

Each string literal is a reference (§4.3) to an instance (§4.3.1, §12.5) of class String (§4.3.3). String objects have a constant value. String literals-or, more generally, strings that are the values of constant expressions (§15.28)-are "interned" so as to share unique instances, using the method String.intern.

所以在 s2 += "d" 的情况下，编译器没有你那么聪明，只是汇集了 "d"。