String s1==String s2（true）但FieldOffset不同

Question

static String s1 = "a";
static String s2 = "a";
System.out.println(s1 == s2);  

http://javapapers.com/core-java/address-of-a-java-object/

但地址好像不一样。我错过了什么？

import sun.misc.Unsafe;
import java.lang.reflect.Field;
public class SomeClass {
    static String s1 = "a";
    static String s2 = "a";
    public static void main (String args[]) throws Exception {
        System.out.println(s1 == s2); //true

        Unsafe unsafe = getUnsafeInstance();
        Field s1Field = SomeClass.class.getDeclaredField("s1");
        System.out.println(unsafe.staticFieldOffset(s1Field)); //600

        Field s2Field = SomeClass.class.getDeclaredField("s2");
        System.out.println(unsafe.staticFieldOffset(s2Field)); //604

    }

    private static Unsafe getUnsafeInstance() throws SecurityException, 
        NoSuchFieldException, IllegalArgumentException, IllegalAccessException {
        Field theUnsafeInstance = Unsafe.class.getDeclaredField("theUnsafe");
        theUnsafeInstance.setAccessible(true);
        return (Unsafe) theUnsafeInstance.get(Unsafe.class);
    }
}

Answer 1

你没有错过任何东西。不安全库正在报告实际发生的情况。

字节码：

static {};
  Code:
   0:   ldc #11; //String a
   2:   putstatic   #13; //Field s1:Ljava/lang/String;
   5:   ldc #11; //String a
   7:   putstatic   #15; //Field s2:Ljava/lang/String;
   10:  return

请注意，这两个字符串都放在内存中的不同位置，即13和15。

变量存储在内存中的位置（需要单独的地址）和新对象是否放在堆上是有区别的。在这种情况下，它为两个变量分配了两个单独的地址，但不需要创建新的String对象，因为它识别相同的String文字。因此这两个变量都引用了相同的字符串。