我在不可变字符串上写了下面的代码。
public class ImmutableStrings {
public static void main(String[] args) {
testmethod();
}
private static void testmethod() {
String a = "a";
System.out.println("a 1-->" + a);
a = "ty";
System.out.println("a 2-->" + a);
}
}
输出:
a 1-->a
a 2-->ty
这里变量a的值被改变了(尽管许多人说不可变对象的内容不能被改变)。但是说字符串是不可变的到底是什么意思呢?你能给我解释一下这个话题吗?
来源:https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
当前回答
If some object bar holds a reference to a mutable object foo and encapsulates some of its state in mutable aspects of foo's state, that will allow code which can change those aspects of foo to change the corresponding aspects of bar's state without actually touching bar or even knowing of its existence. Generally, this means that objects which encapsulate their own state using mutable objects must ensure that no references to those objects are exposed to any code which might unexpectedly mutate them. By contrast, if bar holds a reference to an object moo and only uses immutable aspects of moo other than identity to encapsulate its state, then bar can freely expose moo to outside code without worrying about anything the outside code might do to it.
其他回答
If some object bar holds a reference to a mutable object foo and encapsulates some of its state in mutable aspects of foo's state, that will allow code which can change those aspects of foo to change the corresponding aspects of bar's state without actually touching bar or even knowing of its existence. Generally, this means that objects which encapsulate their own state using mutable objects must ensure that no references to those objects are exposed to any code which might unexpectedly mutate them. By contrast, if bar holds a reference to an object moo and only uses immutable aspects of moo other than identity to encapsulate its state, then bar can freely expose moo to outside code without worrying about anything the outside code might do to it.
你改变了a指的东西。试试这个:
String a="a";
System.out.println("a 1-->"+a);
String b=a;
a="ty";
System.out.println("a 2-->"+a);
System.out.println("b -->"+b);
您将看到a和b所指向的对象没有改变。
如果你想阻止你的代码改变a引用的对象，试试:
final String a="a";
不可变对象是创建后状态不能修改的对象。
a = "ABC" <——immutable object。"a"保存对对象的引用。 并且，a = "DEF" <——另一个不可变对象，"a"现在保存着对它的引用。
一旦你分配了一个字符串对象，该对象就不能在内存中被改变。
总之，您所做的就是将“a”的引用更改为一个新的字符串对象。
只有参照物在改变。首先a引用的是字符串"a"，后来你把它改成了"ty"。字符串“a”保持不变。
字符串是一个char[]，包含一系列UTF-16代码单元，该数组的int偏移量和int长度。
为例。
String s
它为字符串引用创建空间。赋值复制周围的引用，但不修改这些引用所引用的对象。
你也应该意识到
new String(s)
并没有做任何有用的事情。它只是创建了另一个实例，支持的数组、偏移量和长度与s相同。很少有理由这样做，因此大多数Java程序员认为这是一种糟糕的做法。
Java双引号字符串，如“my string”，实际上是对内部string实例的引用，因此“bar”是对同一个string实例的引用，而不管它在代码中出现了多少次。
“hello”创建了一个被池化的实例，而新的String(…)创建了一个非池化的实例。尝试system . out。println((“hello”= =“你好 ") + "," + ( 新的字符串“hello”= =“你好 ") + "," + ( 新的字符串(“hello”)= =新的字符串(“hello”)));你应该看到真，假，假
