Unboxing int value using Integer Class

Question

In this circumstance what is the value of variable y after the first two statements? I'm assuming it's Integer 7 but my book says automatic unboxing of objects only occurs with relational operators < >". I'm a little confused how variable Integer y gets it's value. Does any unboxing occur in newInteger(x)?

Integer x = 7;
Integer y = new Integer(x); 

println( "x == y" + " is " +  (x == y))

possible duplicate of [What exactly does comparing Integers with == do?](http://stackoverflow.com/questions/3689745/what-exactly-does-comparing-integers-with-do) — Abimaran Kugathasan, Jul 13 '15 at 10:32
@AbimaranKugathasan i understand what happens when == compares two objects but I don't understand how `y` gets its value. — Ian, Jul 13 '15 at 10:34
Note that `"x == y" + " is " + x == y` means: `("x == y" + " is " + x) == y`, which is not what you want. Use parentheses: `"x == y" + " is " + (x == y)` — Jesper, Jul 13 '15 at 10:35

OldCurmudgeon · Answer 1 · 2015-07-13T10:51:13.007

Unboxing happens when the compiler is certain that you wish to compare values. Using == can compare the Objects and therefore give false because == is a valid operation between objects. With < and > there is no concept of Object < OtherObject so it is certain that you mean numerically.

public void test() {
    Integer x = 7;
    Integer y = new Integer(x) + 1;

    System.out.println("x == y" + " is " + (x == y));
    System.out.println("x.intValue() == y.intValue()" + " is " + (x.intValue() == y.intValue()));
    System.out.println("x < y" + " is " + (x < y));
    System.out.println("x.intValue() < y.intValue()" + " is " + (x.intValue() < y.intValue()));
}

x == y is false

x.intValue() == y.intValue() is true

x < y is true

x.intValue() < y.intValue() is true

In this circumstance what is the value of variable y after the first two statements?

The value of the variable y is a reference to an Integer object containing the value 7.

score 2 · Accepted Answer · answered Jul 13 '15 at 10:30

2

Integer x = 7;

In this case, the int literal 7 is automatically boxed into the Integer variable x.

Integer y = new Integer(x);

This involves an automatic unboxing of the Integer variable x into an int (temporary) variable, which is passed to the Integer constructor. In other words, it is equivalent to:

Integer y = new Integer(x.intValue());

After this statement, y points to a new object that is different than x but containing the same int wrapped value.

answered Jul 13 '15 at 10:30

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So x in `new Integer(x)` is temporarily unboxed to value type `int` so the integer constructor for variable y could take effect. The sequence of boxing and unboxing was a little confusing to me at first. Now it makes sense. Thanks. – Ian Jul 13 '15 at 10:50
@Ian Yes it does unboxing because the constructor takes a primitive `int`, not the `Integer`. – M A Jul 13 '15 at 11:08

Unboxing int value using Integer Class

2 Answers2