CODEWITHSUNDEEP
javavariablespass-by-reference
Mazzy
Mazzy

Reputation: 1944

Java, how to pass by reference

Have a look at this code

Integer x = 5;
Integer y = 2;
Integer xy = x+y;
System.out.println("xy = " + xy); // outputs: 7
System.out.println("xy2 = " + xy2); // outputs: 7

x++;

System.out.println("xy = " + xy); // outputs: 7
System.out.println("xy2 = " + xy2); // outputs: 7

How can I get the code to output 8 without using a method that calculates it for you?

Upvotes: 2

Views: 150

Answers (4)

Brian Roach
Brian Roach

Reputation: 76918

An Integer in Java is immutable. You can not change its value. In addition, it's a special autoboxing type to provide an Object wrapper for the int primitive.

In your code, for example, x++ does not modify the Integer object x is referencing. It un-autoboxes it to a primitive int, post-increments it, re-autoboxes it returning a new Integer object and assigns that Integer to x.

Edit to add for completeness: Autoboxing is one of those special things in Java that can cause confusion. There's even more going on behind the scenes when talking about memory / objects. The Integer type also implements the flyweight pattern when autoboxing. Values from -128 to 127 are cached. You should always use the .equals() method when comparing Integer objects. 

Integer x = 5;
Integer y = 5;
if (x == y) // == compares the *reference (pointer) value* not the contained int value 
{
    System.out.println("They point to the same object");
}

x = 500;
y = 500;
if (x != y)
{
    System.out.println("They don't point to the same object");
    if (x.equals(y)) // Compares the contained int value
    {
        System.out.println("But they have the same value!");
    }
}

See: Why aren't Integers cached in Java? for more info (and of course the JLS)

Upvotes: 5

Gari BN
Gari BN

Reputation: 1683

The expression: xy = x + y doesn't mean that now xy is depends on the values of x and y (if they changed, xy is changed too). You can see it as follows: the value of the expression x + y is inserted into xy.

Therefore, you must increase the value of x (by x++), before your set the value of xy.

Upvotes: 1

kidcuber
kidcuber

Reputation: 51

I'm new to java, and I'm not really sure of the context for this question, but if all you want to do is output 8, you can just make it xy++ instead of x++.

Integer x = 5;
Integer y = 2;
Integer xy = x+y;
int xy2 = x+y; // just testing to see if it makes a difference
System.out.println("xy = " + xy); // outputs: 7
System.out.println("xy2 = " + xy2); // outputs: 7

**xy++;**
System.out.println("xy = " + xy); // **outputs: 8**
System.out.println("xy2 = " + xy2); // outputs: 7

Upvotes: -1

Simeon Visser
Simeon Visser

Reputation: 122536

You need to update xy after modifying x. So:

x++;
xy = x + y;
xy2 = x + y;

In Java you'll need to update a variable yourself if you want the value to change. It's not like other languages where you express a relationship between two variables and this relationship is maintained whenever a variable changes.

Upvotes: 3

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