Using == operator in Java to compare wrapper objects

integer equals java operator-keyword wrapper

2022-08-31 16:08:55

I'm reading SCJP Java 6 by Kathy Sierra and Bert Bates and this book is confusing me so much. On page 245 they state that the following code below.

Integer i1 = 1000;
Integer i2 = 1000;
if(i1 != i2)
System.out.println("different objects");

//Prints output
different objects

Then on the very next page they have the following code

Integer i3 = 10;
Integer i4 = 10;
if(i3 == i4)
System.out.println("same objects");

//Prints output
same objects

I'm so confused! When I try this out on my own it seems that you cannot use the == to compare the same way you would use equals() method. Using the == always gives me 'false' even if the Integer variables are set to the same value (i.e. 10). Am I correct? Using the == to compare the same Integer object (with same values) will always result in 'false'

答案 1

The key to the answer is called object interning. Java interns small numbers (less than 128), so all instances of with in the interned range are the same. Numbers greater than or equal to 128 are not interned, hence objects are not equal to each other.Integer(n)nInteger(1000)

答案 2

If you look at the source code for you'll see that pools all values -128 to 127. The reason is that small Integer values are used frequently and are thus worthy of being pooled/cached.IntegerInteger.valueOf(int)

Taken straight from :Integer.java

public static Integer valueOf(int i) {
    if(i >= -128 && i <= IntegerCache.high)
        return IntegerCache.cache[i + 128];
    else
        return new Integer(i);
}

Note that this pooling is implementation specific and there's no guarantee of the pooled range.

The answers about interning are correct in concept, but incorrect with terminology. Interning in Java normally implies that the Java runtime is performing the pooling (such as String's intern). In Integer's case it's the class itself that is doing the pooling. There's no JVM magic involved.