“不可变”意味着你不能改变值。如果你有一个String类的实例，你调用的任何方法，只要看起来是在修改这个值，实际上都会创建另一个String。

String foo = "Hello";
foo.substring(3);
<-- foo here still has the same value "Hello"

要保存更改，您应该这样做 Foo = Foo .sustring(3);

当您使用集合时，不可变与可变可能会很有趣。想想如果使用可变对象作为map的键，然后更改值会发生什么(提示:考虑equals和hashCode)。

不可变的对象在创建后不能改变其状态。

尽可能使用不可变对象有三个主要原因，所有这些都将有助于减少你在代码中引入的错误数量:

It is much easier to reason about how your program works when you know that an object's state cannot be changed by another method Immutable objects are automatically thread safe (assuming they are published safely) so will never be the cause of those hard-to-pin-down multithreading bugs Immutable objects will always have the same Hash code, so they can be used as the keys in a HashMap (or similar). If the hash code of an element in a hash table was to change, the table entry would then effectively be lost, since attempts to find it in the table would end up looking in the wrong place. This is the main reason that String objects are immutable - they are frequently used as HashMap keys.

当你知道一个对象的状态是不可变的时，你还可以在代码中做一些其他的优化——例如缓存计算的哈希——但这些都是优化，因此没有那么有趣。

String s1="Hi";
String s2=s1;
s1="Bye";

System.out.println(s2); //Hi  (if String was mutable output would be: Bye)
System.out.println(s1); //Bye

s1="Hi":创建一个包含"Hi"值的对象s1。

S2 =s1:引用s1对象创建对象S2。

s1="Bye":之前的s1对象的值不会改变，因为s1有String类型，而String类型是一个不可变的类型，相反，编译器创建了一个新的String对象与"Bye"值和s1引用它。在这里，当我们打印s2值时，结果将是“Hi”而不是“Bye”，因为s2引用了前一个具有“Hi”值的s1对象。

java.time

这可能有点晚了，但为了理解什么是不可变对象，请考虑以下来自新的Java 8日期和时间API (Java . Time)的示例。你可能知道，Java 8中的所有日期对象都是不可变的，所以在下面的例子中

LocalDate date = LocalDate.of(2014, 3, 18); 
date.plusYears(2);
System.out.println(date);

输出:

2014-03-18

这将打印与初始日期相同的年份，因为plusYears(2)返回一个新对象，因此旧日期仍然不变，因为它是一个不可变对象。一旦创建，您就不能进一步修改它，日期变量仍然指向它。

因此，该代码示例应该捕获并使用由plusYears调用实例化并返回的新对象。

LocalDate date = LocalDate.of(2014, 3, 18); 
LocalDate dateAfterTwoYears = date.plusYears(2);

date.toString()…2014-03-18 dateAfterTwoYears.toString()…2016-03-18

不可变对象是指创建后不能修改的对象。一个典型的例子是字符串字面量。

越来越流行的D编程语言通过“不变”关键字具有“不变性”的概念。查看Dr.Dobb关于它的文章http://dobbscodetalk.com/index.php?option=com_myblog&show=Invariant-Strings.html&Itemid=29。它完美地解释了这个问题。

In large applications its common for string literals to occupy large bits of memory. So to efficiently handle the memory, the JVM allocates an area called "String constant pool".(Note that in memory even an unreferenced String carries around a char[], an int for its length, and another for its hashCode. For a number, by contrast, a maximum of eight immediate bytes is required) When complier comes across a String literal it checks the pool to see if there is an identical literal already present. And if one is found, the reference to the new literal is directed to the existing String, and no new 'String literal object' is created(the existing String simply gets an additional reference). Hence : String mutability saves memory... But when any of the variables change value, Actually - it's only their reference that's changed, not the value in memory(hence it will not affect the other variables referencing it) as seen below....

字符串s1 = "旧字符串";

//s1 variable, refers to string in memory
        reference                 |     MEMORY       |
        variables                 |                  |

           [s1]   --------------->|   "Old String"   |

字符串s2 = s1;

//s2 refers to same string as s1
                                  |                  |
           [s1]   --------------->|   "Old String"   |
           [s2]   ------------------------^

s1 = "New String";

//s1 deletes reference to old string and points to the newly created one
           [s1]   -----|--------->|   "New String"   |
                       |          |                  |
                       |~~~~~~~~~X|   "Old String"   |
           [s2]   ------------------------^

原来的字符串'in memory'没有改变，但是 引用变量已被更改，以便它引用新字符串。 如果我们没有s2, Old String仍然在内存中，但是 我们无法访问它…