字符串串联如何在下面工作 [英] How is String concatenation working in following

问题描述

按照以下> String串联时Java String池如何工作?对话，

String a = "hello world!";
String b = "hello" + " world!";
boolean compare = (a == b);

compare 应该为true，这是正确的.但是，我有以下代码

The compare should be true which is correct. However, I've the following code

String s1 = "This is";
String s2 = " a new String";
String s3 = s1 + s2;
String s4 = "This is a new String";

在比较 System.out.printf("s3 == s4:％s \ n"，s3 == s4);//始终为假这总是错误的.我的理解是，在 s1 & s2 ， s3 将在池中创建一个字符串，并且在创建 s4 时将指向相同的池位置.但这种情况并非如此.我已经尝试过使用包括7、8和14在内的其他JDK进行此操作，结果是一致的.

On comparing System.out.printf("s3 == s4:%s\n", s3 == s4);//always false It is always false. My understanding is that on concatenation of s1 & s2, s3 will create a string in pool and when s4 is created it will point to the same pool location. But this is not the case. I've tried this with different JDKs including 7, 8 and 14 and the results are consistent.

推荐答案

这是在Java语言规范(重点是我的)中指定有关池字符串的行为的方法:

This is how the behaviour regarding pooling strings is specified in the Java Language Specification (emphasis mine):

3.10.5字符串文字

[...]

此外，字符串文字总是引用类 String 的相同实例.这是因为字符串文字-或更一般地说，是常量表达式值的 字符串 (第15.28节)-是"interned"的.以便使用方法 String.intern 共享唯一的实例.

Moreover, a string literal always refers to the same instance of class String. This is because 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.

仅常量表达式 String 的a>.这是一个新的字符串" ，以及"hello"+"世界！" 是常量表达式.编译器可以在编译时评估这些表达式. s1 + s2 不是常数表达式.

Only constant expressions of type String are pooled. "This is a new String", and "hello" + " world!" are constant expressions. The compiler can evaluate these expressions at compile time. s1 + s2 is not a constant expression.

因此，当执行 s1 + s2 时，会创建一个新的字符串 .但是请注意，另一个字符串恰好与新字符串具有相同的字符，它位于字符串池中，因为您使用了字符串文字来初始化 s4 .

So when executing s1 + s2, a new string is created. But note that another string, which just so happens to have the same characters as the new string, is in the string pool, because you used a string literal to initialise s4.

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