无法理解 Java 字符串文字是如何实现的

Question

我正在学习 StringBuider class。我在这个网站和许多其他书籍中读到，当编译器遇到文字的“+”运算符时，它会自动使用 StringBuilder append 的方法来连接它们。

这似乎有点问题，因为 StringBuilder 对象将在运行时创建，但 String 引用应该在编译时就已获取连接 String 对象的地址。

String s1 = "hello";  
String s2 ="bc"; 
int value = 22;

编译时"meets"这段代码：

String s = s1+s2+22;

它"changes"它到：

String s = new StringBuilder().append("hello").append("bc").append(22).toString();

可能是我误会了什么？

Answer 1

15.18.1. String Concatenation Operator +

An implementation may choose to perform conversion and concatenation in one step to avoid creating and then discarding an intermediate String object. To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.

在你的情况下，

String s1 = "hello";
String s2 ="bc";
int value = 22;

String r = s1 + s2 + value;

你会得到

INVOKESPECIAL java/lang/StringBuilder.<init> ()V
ALOAD 1
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 2
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ILOAD 3
INVOKEVIRTUAL java/lang/StringBuilder.append (I)Ljava/lang/StringBuilder;
INVOKEVIRTUAL java/lang/StringBuilder.toString ()Ljava/lang/String;

连接常量对象时

String r = "hello" + "bc" + 22;

你会得到

LDC "hellobc22"
ASTORE 2

When the compiler "meets" this code:
String s = s1+s2+22;
it "changes" it to:
String s = new StringBuilder().append("hello").append("bc").append(22).toString();

没有。它可能会将其优化为

String s = new StringBuilder().append(s1).append(s2).append(value).toString();

但它不能用 "hello" 替换 s1，因为不能保证变量会一直引用 "hello"。该变量不是最终变量，因此可以重新分配。

无法理解 Java 字符串文字是如何实现的

Can't understand how Java string literals are implemented

java

string

jvm

compiler-optimization

stringbuilder

15.18.1. String Concatenation Operator +