Are Java wrapper classes really immutable?

后端 未结 6 1681
借酒劲吻你
借酒劲吻你 2020-12-16 12:16

Java Wrapper classes are supposed to be immutable. This means that once an object is being created, e.g.,

Integer i = new Integer(5);

its

相关标签:
6条回答
  • 2020-12-16 12:32
    Integer i = new Integer(5);
    i++;                         // i  will become 6
    

    where i++ is the same with i = new Integer( i.intValue() + 1);

    0 讨论(0)
  • 2020-12-16 12:50

    The reason i = 6 works is that auto-boxing is intercepting and turning it into i = new Integer(6). Thus as @Peter said, you are now pointing at a new object.

    0 讨论(0)
  • 2020-12-16 12:51

    Immutable means that the object state cannot be changed. In your case you haven't changed the object new Integer(5), but you have changed the reference i to point to another object. Hope it is clear:)

    0 讨论(0)
  • 2020-12-16 12:52

    All wrapper classes in java are immutable. We can't change the value of a wrapper class object once created, i.e., can't change the value wrapped inside the object. Because wrapper classes are used as object form of primitive data types and if they are mutable, data inconsistencies will occur in runtime. However, you can change the wrapper class reference variable to hold another object.

    0 讨论(0)
  • 2020-12-16 12:53

    i is a reference. Your code change the reference i to point to a different, equally immutable, Integer.

    final Integer i = Integer.valueOf(5);
    

    might be more useful.

    0 讨论(0)
  • 2020-12-16 12:54

    The compiler autoboxes primitive values, this means that

    Integer value = 6;
    

    will be compiled as

    Integer value = Integer.valueOf(6);
    

    Integer.valueOf will return an Integer instance with the given value. In your case i will now reference the Integer(6) instead of the Integer(5), the Integer(5) object itself will not change.

    To see this you can do following

    Integer i = new Integer(5);//assign new integer to i
    Integer b = i;//b refences same integer as i
    i = 6;//modify i
    System.out.println(i +"!="+b);
    

    This will print 6!=5, if the integer instance had been modified it would print 6!=6 instead.

    To clarify this is only meant to show how an assignment to Integer only modifies the reference and does not alter the Integer instance itself. As user @KNU points out it does not prove or show the immutability of Integer, as far as I can tell the immutability is only indirectly given by the lack of modifying methods in its API and the requirement that instances returned by Integer.valueOf have to be cached for a certain range.

    0 讨论(0)
提交回复
热议问题