理解String

Stringstr=”kvill;
Stringstr=newString(“kvill”);的区别：

Strings0=”kvill”;
Strings1=”kvill”;
Strings2=”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：

packagetestPackage;
classTest...{
publicstaticvoidmain(String[]args)...{
Stringhello="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());
}
}
classOther...{staticStringhello="Hello";}

packageother;
publicclassOther...{staticStringhello="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.