Java 中如何连接最终字符串？

Question

当我编译这段代码时。

public class InternTest {
    public static void main(String...strings ){
        final String str1="str";
        final String str2="ing";
        String str= str1+str2;

    }
}

产生以下字节码

public static void main(java.lang.String...);
   flags: ACC_PUBLIC, ACC_STATIC, ACC_VARARGS
   Code:
     stack=1, locals=4, args_size=1
        0: ldc           #16                 // String str
        2: astore_1
        3: ldc           #18                 // String ing
        5: astore_2
        6: ldc           #20                 // String string
        8: astore_3
        9: return

所以字符串文字 "string" 已经存在于常量池中，并被推送 6: ldc #20 // String string 到栈中这一行。

引用 JSL

From JLS §4.12.4 - final Variables:

A variable of primitive type or type String, that is final and initialized with a compile-time constant expression (§15.28), is called a constant variable.

Also from JLS §15.28 - ConstantExpression:

Compile-time constant expressions of type String are always "interned" so as to share unique instances, using the method String#intern().

所以我知道 str1 和 str2 将在创建时被保留。"str" 和 "ing" 将在第 String str= str1+str2; 行共享相同的内存 但是str1+str2如何在常量字符串池中直接产生"string"。没有像我不写 final. 时那样调用任何 String Builder class ？看看是不是跟实习生的事情有关系

我写了这个片段

public class IntermTest {
    public static void main(String...strings ){
         String str1=("str").intern();
        String str2=("ing").intern();
        String str= str1+str2;

    }
}

但是当我生成字节码时，我得到了这个

public static void main(java.lang.String...);
    flags: ACC_PUBLIC, ACC_STATIC, ACC_VARARGS
    Code:
      stack=3, locals=4, args_size=1
         0: ldc           #16                 // String str
         2: invokevirtual #18                 // Method java/lang/String.intern:
()Ljava/lang/String;
         5: astore_1
         6: ldc           #24                 // String ing
         8: invokevirtual #18                 // Method java/lang/String.intern:
()Ljava/lang/String;
        11: astore_2
        12: new           #26                 // class java/lang/StringBuilder
        15: dup
        16: aload_1
        17: invokestatic  #28                 // Method java/lang/String.valueOf
:(Ljava/lang/Object;)Ljava/lang/String;
        20: invokespecial #32                 // Method java/lang/StringBuilder.
"<init>":(Ljava/lang/String;)V
        23: aload_2
        24: invokevirtual #35                 // Method java/lang/StringBuilder.
append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
        27: invokevirtual #39                 // Method java/lang/StringBuilder.
toString:()Ljava/lang/String;
        30: astore_3
        31: return

实际上它也使用 stringBuilder 进行连接。所以它对final做了一些事情。 final 字符串有什么我绝对不知道的特别之处吗？

Answer 1

http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.28 表示

Simple names (§6.5.6.1) that refer to constant variables (§4.12.4) are constant expressions.

http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.28 还说：

A constant expression is an expression denoting a value of primitive type or a String that does not complete abruptly and is composed using only the following:

Literals of primitive type and literals of type String (§3.10.1, §3.10.2, §3.10.3, §3.10.4, §3.10.5)

[...]

The additive operators + and - (§15.18)

[...]

Simple names (§6.5.6.1) that refer to constant variables (§4.12.4).

Example 15.28-1. Constant Expressions

[...]

"The integer " + Long.MAX_VALUE + " is mighty big."

由于这两个变量是常量表达式，编译器会进行连接：

String str = str1 + str2;

的编译方式与

相同

String str = "str" + "ing";

编译方式与

相同

String str = "string";

Java 中如何连接最终字符串？

How is concatenation of final strings done in Java?

java

string

final