Integer Wrapper Objects Share the Same Instances Only Within the Value 127

Why is 128==128 false but 127==127 is true when comparing Integer wrappers in Java?

When you compile a number literal in Java and assign it to a Integer (capital I) the compiler emits:

Integer b2 =Integer.valueOf(127)

This line of code is also generated when you use autoboxing.

valueOf is implemented such that certain numbers are "pooled", and it returns the same instance for values smaller than 128.

From the java 1.6 source code, line 621:

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

The value of high can be configured to another value, with the system property.

-Djava.lang.Integer.IntegerCache.high=999

If you run your program with that system property, it will output true!

The obvious conclusion: never rely on two references being identical, always compare them with .equals() method.

So b2.equals(b3) will print true for all logically equal values of b2,b3.

Note that Integer cache is not there for performance reasons, but rather to conform to the JLS, section 5.1.7; object identity must be given for values -128 to 127 inclusive.

Integer#valueOf(int) also documents this behavior:

this method is likely to yield significantly better space and time performance by caching frequently requested values. This method will always cache values in the range -128 to 127, inclusive, and may cache other values outside of this range.

java == for Integer

Since Java 5, wrapper class caching was introduced. The following is an examination of the cache created by an inner class, IntegerCache, located in the Integer cache. For example, the following code will create a cache:

Integer myNumber = 10

or

Integer myNumber = Integer.valueOf(10);

256 Integer objects are created in the range of -128 to 127 which are all stored in an Integer array. This caching functionality can be seen by looking at the inner class, IntegerCache, which is found in Integer:

So when creating an object using Integer.valueOf or directly assigning a value to an Integer within the range of -128 to 127 the same object will be returned. Therefore, consider the following example:

Integer i = 100;
Integer p = 100;
if (i == p)  
    System.out.println("i and p are the same.");
if (i != p)
    System.out.println("i and p are different.");   
if(i.equals(p))
    System.out.println("i and p contain the same value.");

The output is:

i and p are the same.
i and p contain the same value.

It is important to note that object i and p only equate to true because they are the same object, the comparison is not based on the value, it is based on object equality. If Integer i and p are outside the range of -128 or 127 the cache is not used, therefore new objects are created. When doing a comparison for value always use the “.equals” method. It is also important to note that instantiating an Integer does not create this caching.

Remember that “==” is always used for object equality, it has not been overloaded for comparing unboxed values

Using == operator in Java to compare wrapper objects

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

java Map with Integer value comparison

This is because it's caching by JVM. In other words for all values [-128; 128). Integer.valueOf(i) will return the same Integer instance.

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

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