理解String

 String str=”kvill; 
String str=new String (“kvill”);的区别：  

    String s0=”kvill”; 
    String s1=”kvill”; 
    String s2=”kv” + “ill”; 
    System.out.println( s0==s1 );                     //  结果是  True
    System.out.println( s0==s2 );                     //  结果也是  True

Java会确保一个字符串常量只有一个拷贝。 
因为例子中的s0和s1中的”kvill”都是字符串常量，它们在编译期就被确定了，所以s0==s1为true；而”kv”和”ill”也都是字符串常量，当一个字符串由多个字符串常量连接而成时，它自己肯定也是字符串常量，所以 s2也同样在编译期就被解析为一个字符串常量，所以s2也是常量池中”kvill”的一个引用。 
所以我们得出s0==s1==s2; 

在JAVA language specification中的3.10.5 String Literals有相关说明：

the test program consisting of the compilation unit：

package testPackage;
class Test ...{
   public static void main(String[] args) ...{
        String hello = "Hello", lo = "lo";
        System.out.print((hello == "Hello") + " ");
        System.out.print((Other.hello == hello) + " ");
        System.out.print((other.Other.hello == hello) + " ");
        System.out.print((hello == ("Hel"+"lo")) + " ");
        System.out.print((hello == ("Hel"+lo)) + " ");
        System.out.println(hello == ("Hel"+lo).intern());
        }
}
class Other ...{ static String hello = "Hello"; }

package other;
public class Other ...{ static String hello = "Hello"; }

produces the output:

                                     true true true true false true

This example illustrates six points:
• Literal strings within the same class in the same package represent
references to the same String object .
• Literal strings within different classes in the same package represent references
to the same String object.
• Literal strings within different classes in different packages likewise represent
references to the same String object.
• Strings computed by constant expressions  are computed at compile
time and then treated as if they were literals.
• Strings computed by concatenation at run time are newly created and therefore
distinct.
The result of explicitly interning a computed string is the same string as any
pre-existing literal string with the same contents.