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))

Answer 1

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

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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.