Why Does Java's `==` Operator Sometimes Compare Integer Object Values and Sometimes References?

Surprising Integer Boxing Behavior in Java

In Java, an Integer object is a wrapper for a primitive int value. When boxing an int into an Integer, one would typically expect the == operator to compare the values of the objects. However, in certain cases, the == operator instead compares the references of the objects. This behavior can lead to surprising results.

Consider the following code:

public class Scratch {
    public static void main(String[] args) {
        Integer a = 1000, b = 1000;
        System.out.println(a == b);

        Integer c = 100, d = 100;
        System.out.println(c == d);
    }
}

When run, this code prints:

false
true

The first output is as expected: == compares the references of two distinct Integer objects. However, the second output is surprising. Why does == return true when c and d have the same value?

The answer lies in the Java Language Specification (JLS). According to Section 5.1.7 of the JLS:

If the value p being boxed is true, false, a byte, a char in the range u0000 to u007f, or an int or short number between -128 and 127, then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.

This means that Integer objects created from primitive values within the specified ranges are guaranteed to have identical references. In our case, c and d are both created from the same int value, 100, which falls within the range of -128 to 127. Therefore, they have the same reference, and the == operator returns true.

While the behavior of the second line of output is guaranteed, the JLS suggests that the behavior of the first line of output is not. In theory, a Java implementation could cache Integer objects for common values to improve performance, but the JLS does not require this behavior. As a result, different Java implementations may handle this case differently.

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