Question

以下内容如何返回“ false ”？

    String a = "123";
    a = a + "456";
    String b = "123456";
    System.out.println((a == b));

据我了解，

字符串“123”在字符串池中创建并分配给“a”
在池中创建字符串“456”，并在池中创建“123456”，“a”开始引用它。
为值“123456”创建“b”时; JVM会在字符串池中发现现有的字符串“123456”，而“b”也会引用它。

因此它应该返回true！

我哪里错了？

Answer 1

这一行：a = a + "456";将在堆中创建一个新对象（你正在连接）并将其分配给a，这就是你弄错的原因。您可以调用intern方法（将字符串从堆放到池中）：a.intern() == b然后它将是true。

Answer 2

在您的示例中，

String a = "123"; //a reference to "123" in string pool
a = a + "456"; // Here as you are concatenating using + operator which create a new String object in heap and now a is referencing a string object in heap instead of a string literal in string pool. 
String b = "123456"; // here b is referencing to string pool for "123456"
System.out.println((a == b)); // This will return false because for the value "123456" a is referencing to heap and b to string pool. Because == operator compare reference rather then values it will return false.

For more details you can read this page

连接String和整个String之间的比较返回false;

2 个答案: