How can I pass an Integer class correctly by reference

Question

I am hoping that someone can clarify what is happening here for me   I dug around in the integer class for a bit but because integer is overriding the   operator I could not figure out what was going wrong   My problem is with this line   Integer i   0  i   i   1        I think that this is somehow creating a new object    Here is my reasoning   I know that java is pass by value  or pass by value of reference   so I think that in the following example the integer object should be incremented each time     public class PassByReference        public static Integer inc Integer i            i   i 1        I think that this must be   sneakally   creating a new integer              System out println  Inc    i           return i             public static void main String   args            Integer integer   new Integer 0           for  int i  0  i lt 10  i                 inc integer               System out println  main    integer                       This is my expected output    Inc  1 main  1 Inc  2 main  2 Inc  3 main  3 Inc  4 main  4 Inc  5 main  5 Inc  6 main  6       This is the actual output      Inc  1 main  0 Inc  1 main  0 Inc  1 main  0       Why is it behaving like this

User · Accepted Answer

There are two problems    Integer is pass by value  not by reference  Changing the reference inside a method won t be reflected into the passed-in reference in the calling method  Integer is immutable  There s no such method like Integer set i   You could otherwise just make use of it    To get it to work  you need to reassign the return value of the inc   method   integer   inc integer       To learn a bit more about passing by value  here s another example   public static void main String    args        String   strings   new String      foo    bar         changeReference strings       System out println Arrays toString strings       still  foo  bar      changeValue strings       System out println Arrays toString strings        foo  foo    public static void changeReference String   strings        strings   new String      foo    foo       public static void changeValue String   strings        strings 1     foo

User · Answer

What you are seeing here is not an overloaded   oparator  but autoboxing behaviour  The Integer class is immutable and your code   Integer i   0  i   i   1      is seen by the compiler  after the autoboxing  as   Integer i   Integer valueOf 0   i   Integer valueOf i intValue     1       so you are correct in your conclusion that the Integer instance is changed  but not sneakily - it is consistent with the Java language definition  -

User · Answer

The Integer is immutable  You can wrap int in your custom wrapper class   class WrapInt      int value     WrapInt theInt   new WrapInt     inc theInt   System out println  main    theInt value

User · Answer

You are correct here   Integer i   0  i   i   1       lt - I think that this is somehow creating a new object    First  Integer is immutable     Second  the Integer class is not overriding the   operator  there is autounboxing and autoboxing involved at that line  In older versions of Java you would get an error on the above line   When you write i   1 the compiler first converts the Integer to an  primitive  int for performing the addition  autounboxing  Next  doing i    lt some int gt  the compiler converts from int to an  new  Integer  autoboxing  So   is actually being applied to primitive ints

User · Answer

Good answers above explaining the actual question from the OP    If anyone needs to pass around a number that needs to be globally updated  use the AtomicInteger   instead of creating the various wrapper classes suggested or relying on 3rd party libs    The AtomicInteger   is of course mostly used for thread safe access but if the performance hit is no issue  why not use this built-in class  The added bonus is of course the obvious thread safety   import java util concurrent atomic AtomicInteger

User · Answer

I think it is the autoboxing that is throwing you off   This part of your code      public static Integer inc Integer i            i   i 1        I think that this must be   sneakally   creating a new integer              System out println  Inc    i           return i          Really boils down to code that looks like     public static Integer inc Integer i            i   new Integer i    new Integer 1                 System out println  Inc    i           return i          Which of course   will not changes the reference passed in    You could fix it with something like this    public static void main String   args            Integer integer   new Integer 0           for  int i  0  i lt 10  i                 integer   inc integer               System out println  main    integer

User · Answer

If you change your inc   function to this   public static Integer inc Integer i          Integer iParam   i        i   i 1        I think that this must be   sneakally   creating a new integer            System out println i    iParam         return i        then you will see that it always prints  false   That means that the addition creates a new instance of Integer and stores it in the local variable i   local   because i is actually a copy of the reference that was passed   leaving the variable of the calling method untouched   Integer is an immutable class  meaning that you cannot change it s value but must obtain a new instance  In this case you don t have to do it manually like this   i   new Integer i 1     actually  you would use Integer valueOf i intValue   1     instead  it is done by autoboxing

User · Answer

1   Only the copy of reference is sent as a value to the formal parameter  When the formal parameter variable is assigned other value  the formal parameter s reference changes but the actual parameter s reference remain the same incase of this integer object     public class UnderstandingObjects    public static void main String   args         Integer actualParam   new Integer 10        changeValue actualParam        System out println  Output     actualParam      o p  10      IntObj obj   new IntObj         obj setVal 20        changeValue obj        System out println obj a      o p  200     private static void changeValue Integer formalParam         formalParam   100          Only the copy of reference is set to the formal parameter        this is something like   gt  Integer formalParam  new Integer 100          Here we are changing the reference of formalParam itself not just the        reference value     private static void changeValue IntObj obj        obj setVal 200                  obj   new IntObj    obj setVal 200                  Here we are not changing the reference of obj  we are just changing the        reference obj s value         we are not doing obj   new IntObj     obj setValue 200   which has happend        with the Integer         class IntObj       Integer a   public void setVal int a        this a   a

User · Answer

There are 2 ways to pass by reference   Use org apache commons lang mutable MutableInt from Apache Commons library   Create custom class as shown below   Here s a sample code to do it   public class Test       public static void main String args              Integer a   new Integer 1           Integer b   a          Test modify a           System out println a           System out println b            IntegerObj ao   new IntegerObj 1           IntegerObj bo   ao          Test modify ao           System out println ao value           System out println bo value               static void modify Integer x            x 7            static void modify IntegerObj x            x value 7              class IntegerObj       int value      IntegerObj int val            this value   val            Output   1 1 7 7

[java] How can I pass an Integer class correctly by reference?

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