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Equality not working while comparing Integer and int [duplicate]

This question already has answers here:
Why is 128==128 false but 127==127 is true when comparing Integer wrappers in Java?
(8 answers)
Closed 8 years ago.
Why Integer == operator does not work for 128 and after Integer values? Can someone explain this situation?
This is my Java environment:
java version "1.6.0_37"
Java(TM) SE Runtime Environment (build 1.6.0_37-b06)
Java HotSpot(TM) 64-Bit Server VM (build 20.12-b01, mixed mode)
Sample Code:
Integer a;
Integer b;
a = 129;
b = 129;
for (int i = 0; i < 200; i++) {
a = i;
b = i;
if (a != b) {
System.out.println("Value:" + i + " - Different values");
} else {
System.out.println("Value:" + i + " - Same values");
}
}
Some part of console output:
Value:124 - Same values
Value:125 - Same values
Value:126 - Same values
Value:127 - Same values
Value:128 - Different values
Value:129 - Different values
Value:130 - Different values
Value:131 - Different values
Value:132 - Different values

Check out the source code of Integer . You can see the caching of values there.
The caching happens only if you use Integer.valueOf(int), not if you use new Integer(int). The autoboxing used by you uses Integer.valueOf.
According to the JLS, you can always count on the fact that for values between -128 and 127, you get the identical Integer objects after autoboxing, and on some implementations you might get identical objects even for higher values.
Actually in Java 7 (and I think in newer versions of Java 6), the implementation of the IntegerCache class has changed, and the upper bound is no longer hardcoded, but it is configurable via the property "java.lang.Integer.IntegerCache.high", so if you run your program with the VM parameter -Djava.lang.Integer.IntegerCache.high=1000, you get "Same values" for all values.
But the JLS still guarantees it only until 127:
Ideally, boxing a given primitive value p, would always yield an identical reference. In practice, this may not be feasible using existing implementation techniques. The rules above are a pragmatic compromise. The final clause above requires that certain common values always be boxed into indistinguishable objects. The implementation may cache these, lazily or eagerly.
For other values, this formulation disallows any assumptions about the identity of the boxed values on the programmer's part. This would allow (but not require) sharing of some or all of these references.
This ensures that in most common cases, the behavior will be the desired one, without imposing an undue performance penalty, especially on small devices. Less memory-limited implementations might, for example, cache all characters and shorts, as well as integers and longs in the range of -32K - +32K.

Integer is a wrapper class for int.
Integer != Integer compares the actual object reference, where int != int will compare the values.
As already stated, values -128 to 127 are cached, so the same objects are returned for those.
If outside that range, separate objects will be created so the reference will be different.
To fix it:
Make the types int or
Cast the types to int or
Use .equals()

According to Java Language Specifications:
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.
JLS Boxing Conversions
Refer to this article for more information on int caching

The Integer object has an internal cache mechanism:
private static class IntegerCache {
static final int high;
static final Integer cache[];
static {
final int low = -128;
// high value may be configured by property
int h = 127;
if (integerCacheHighPropValue != null) {
// Use Long.decode here to avoid invoking methods that
// require Integer's autoboxing cache to be initialized
int i = Long.decode(integerCacheHighPropValue).intValue();
i = Math.max(i, 127);
// Maximum array size is Integer.MAX_VALUE
h = Math.min(i, Integer.MAX_VALUE - -low);
}
high = h;
cache = new Integer[(high - low) + 1];
int j = low;
for(int k = 0; k < cache.length; k++)
cache[k] = new Integer(j++);
}
private IntegerCache() {}
}
Also see valueOf method:
public static Integer valueOf(int i) {
if(i >= -128 && i <= IntegerCache.high)
return IntegerCache.cache[i + 128];
else
return new Integer(i);
}
This is why you should use valueOf instead of new Integer. Autoboxing uses this cache.
Also see this post: https://effective-java.com/2010/01/java-performance-tuning-with-maximizing-integer-valueofint/
Using == is not a good idea, use equals to compare the values.

Use .equals() instead of ==.
Integer values are only cached for numbers between -127 and 128, because they are used most often.
if (a.equals(b)) { ... }

Depending on how you get your Integer instances, it may not work for any value:
System.out.println(new Integer(1) == new Integer(1));
prints
false
This is because the == operator applied to reference-typed operands has nothing to do with the value those operands represent.

It's because Integer class implementation logic. It has prepared objects for numbers till 128. You can checkout http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/lang/Integer.java source of open-jdk for example (search for cache[]).
Basically objects shouldn't be compared using == at all, with one exception to Enums.

Primitive int vs Integer type [duplicate]

I just saw code similar to this:
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 ran, this block of code will print out:
false
true
I understand why the first is false: because the two objects are separate objects, so the == compares the references. But I can't figure out, why is the second statement returning true? Is there some strange autoboxing rule that kicks in when an Integer's value is in a certain range? What's going on here?

The true line is actually guaranteed by the language specification. From section 5.1.7:
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.
The discussion goes on, suggesting that although your second line of output is guaranteed, the first isn't (see the last paragraph quoted below):
Ideally, boxing a given primitive
value p, would always yield an
identical reference. In practice, this
may not be feasible using existing
implementation techniques. The rules
above are a pragmatic compromise. The
final clause above requires that
certain common values always be boxed
into indistinguishable objects. The
implementation may cache these, lazily
or eagerly.
For other values, this formulation
disallows any assumptions about the
identity of the boxed values on the
programmer's part. This would allow
(but not require) sharing of some or
all of these references.
This ensures that in most common
cases, the behavior will be the
desired one, without imposing an undue
performance penalty, especially on
small devices. Less memory-limited
implementations might, for example,
cache all characters and shorts, as
well as integers and longs in the
range of -32K - +32K.

public class Scratch
{
public static void main(String[] args)
{
Integer a = 1000, b = 1000; //1
System.out.println(a == b);
Integer c = 100, d = 100; //2
System.out.println(c == d);
}
}
Output:
false
true
Yep the first output is produced for comparing reference; 'a' and 'b' - these are two different reference. In point 1, actually two references are created which is similar as -
Integer a = new Integer(1000);
Integer b = new Integer(1000);
The second output is produced because the JVM tries to save memory, when the Integer falls in a range (from -128 to 127). At point 2 no new reference of type Integer is created for 'd'. Instead of creating a new object for the Integer type reference variable 'd', it only assigned with previously created object referenced by 'c'. All of these are done by JVM.
These memory saving rules are not only for Integer. for memory saving purpose, two instances of the following wrapper objects (while created through boxing), will always be == where their primitive values are the same -
Boolean
Byte
Character from \u0000 to \u007f (7f is 127 in decimal)
Short and Integer from -128 to 127

Integer objects in some range (I think maybe -128 through 127) get cached and re-used. Integers outside that range get a new object each time.

Integer Cache is a feature that was introduced in Java Version 5 basically for :
Saving of Memory space
Improvement in performance.
Integer number1 = 127;
Integer number2 = 127;
System.out.println("number1 == number2" + (number1 == number2);
OUTPUT: True
Integer number1 = 128;
Integer number2 = 128;
System.out.println("number1 == number2" + (number1 == number2);
OUTPUT: False
HOW?
Actually when we assign value to an Integer object, it does auto promotion behind the hood.
Integer object = 100;
is actually calling Integer.valueOf() function
Integer object = Integer.valueOf(100);
Nitty-gritty details of valueOf(int)
public static Integer valueOf(int i) {
if (i >= IntegerCache.low && i <= IntegerCache.high)
return IntegerCache.cache[i + (-IntegerCache.low)];
return new Integer(i);
}
Description:
This method will always cache values in the range -128 to 127,
inclusive, and may cache other values outside of this range.
When a value within range of -128 to 127 is required it returns a constant memory location every time.
However, when we need a value thats greater than 127
return new Integer(i);
returns a new reference every time we initiate an object.
Furthermore, == operators in Java is used to compares two memory references and not values.
Object1 located at say 1000 and contains value 6.
Object2 located at say 1020 and contains value 6.
Object1 == Object2 is False as they have different memory locations though contains same values.

Yes, there is a strange autoboxing rule that kicks in when the values are in a certain range. When you assign a constant to an Object variable, nothing in the language definition says a new object must be created. It may reuse an existing object from cache.
In fact, the JVM will usually store a cache of small Integers for this purpose, as well as values such as Boolean.TRUE and Boolean.FALSE.

My guess is that Java keeps a cache of small integers that are already 'boxed' because they are so very common and it saves a heck of a lot of time to re-use an existing object than to create a new one.

That is an interesting point.
In the book Effective Java suggests always to override equals for your own classes. Also that, to check equality for two object instances of a java class always use the equals method.
public class Scratch
{
public static void main(String[] args)
{
Integer a = 1000, b = 1000;
System.out.println(a.equals(b));
Integer c = 100, d = 100;
System.out.println(c.equals(d));
}
}
returns:
true
true

Direct assignment of an int literal to an Integer reference is an example of auto-boxing, where the literal value to object conversion code is handled by the compiler.
So during compilation phase compiler converts Integer a = 1000, b = 1000; to Integer a = Integer.valueOf(1000), b = Integer.valueOf(1000);.
So it is Integer.valueOf() method which actually gives us the integer objects, and if we look at the source code of Integer.valueOf() method we can clearly see the method caches integer objects in the range -128 to 127 (inclusive).
/**
* Returns an {#code Integer} instance representing the specified
* {#code int} value. If a new {#code Integer} instance is not
* required, this method should generally be used in preference to
* the constructor {#link #Integer(int)}, as 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.
*
* #param i an {#code int} value.
* #return an {#code Integer} instance representing {#code i}.
* #since 1.5
*/
public static Integer valueOf(int i) {
if (i >= IntegerCache.low && i <= IntegerCache.high)
return IntegerCache.cache[i + (-IntegerCache.low)];
return new Integer(i);
}
So instead of creating and returning new integer objects, Integer.valueOf() the method returns Integer objects from the internal IntegerCache if the passed int literal is greater than -128 and less than 127.
Java caches these integer objects because this range of integers gets used a lot in day to day programming which indirectly saves some memory.
The cache is initialized on the first usage when the class gets loaded into memory because of the static block. The max range of the cache can be controlled by the -XX:AutoBoxCacheMax JVM option.
This caching behaviour is not applicable for Integer objects only, similar to Integer.IntegerCache we also have ByteCache, ShortCache, LongCache, CharacterCache for Byte, Short, Long, Character respectively.
You can read more on my article Java Integer Cache - Why Integer.valueOf(127) == Integer.valueOf(127) Is True.

In Java the boxing works in the range between -128 and 127 for an Integer. When you are using numbers in this range you can compare it with the == operator. For Integer objects outside the range you have to use equals.

If we check the source code of the Integer class, we can find the source of the valueOf method just like this:
public static Integer valueOf(int i) {
if (i >= IntegerCache.low && i <= IntegerCache.high)
return IntegerCache.cache[i + (-IntegerCache.low)];
return new Integer(i);
}
This explains why Integer objects, which are in the range from -128 (Integer.low) to 127 (Integer.high), are the same referenced objects during the autoboxing. And we can see there is a class IntegerCache that takes care of the Integer cache array, which is a private static inner class of the Integer class.
There is another interesting example that may help to understand this weird situation:
public static void main(String[] args) throws ReflectiveOperationException {
Class cache = Integer.class.getDeclaredClasses()[0];
Field myCache = cache.getDeclaredField("cache");
myCache.setAccessible(true);
Integer[] newCache = (Integer[]) myCache.get(cache);
newCache[132] = newCache[133];
Integer a = 2;
Integer b = a + a;
System.out.printf("%d + %d = %d", a, a, b); // The output is: 2 + 2 = 5
}

In Java 5, a new feature was introduced to save the memory and improve performance for Integer type objects handlings. Integer objects are cached internally and reused via the same referenced objects.
This is applicable for Integer values in range between –127 to +127
(Max Integer value).
This Integer caching works only on autoboxing. Integer objects will
not be cached when they are built using the constructor.
For more detail pls go through below Link:
Integer Cache in Detail

Class Integer contains cache of values between -128 and 127, as it required by JLS 5.1.7. Boxing Conversion. So when you use the == to check the equality of two Integers in this range, you get the same cached value, and if you compare two Integers out of this range, you get two diferent values.
You can increase the cache upper bound by changing the JVM parameters:
-XX:AutoBoxCacheMax=<cache_max_value>
or
-Djava.lang.Integer.IntegerCache.high=<cache_max_value>
See inner IntegerCache class:
/**
* Cache to support the object identity semantics of autoboxing for values
* between -128 and 127 (inclusive) as required by JLS.
*
* The cache is initialized on first usage. The size of the cache
* may be controlled by the {#code -XX:AutoBoxCacheMax=<size>} option.
* During VM initialization, java.lang.Integer.IntegerCache.high property
* may be set and saved in the private system properties in the
* sun.misc.VM class.
*/
private static class IntegerCache {
static final int low = -128;
static final int high;
static final Integer cache[];
static {
// high value may be configured by property
int h = 127;
String integerCacheHighPropValue =
sun.misc.VM.getSavedProperty("java.lang.Integer.IntegerCache.high");
if (integerCacheHighPropValue != null) {
try {
int i = parseInt(integerCacheHighPropValue);
i = Math.max(i, 127);
// Maximum array size is Integer.MAX_VALUE
h = Math.min(i, Integer.MAX_VALUE - (-low) -1);
} catch( NumberFormatException nfe) {
// If the property cannot be parsed into an int, ignore it.
}
}
high = h;
cache = new Integer[(high - low) + 1];
int j = low;
for(int k = 0; k < cache.length; k++)
cache[k] = new Integer(j++);
// range [-128, 127] must be interned (JLS7 5.1.7)
assert IntegerCache.high >= 127;
}
private IntegerCache() {}
}

Java: Why is constant pool maintained only for String values ?

My question is about java interning and constant pools.
Java maintains a a constants pool for java.lang.String, to use JVM memory cleverly, and to do so java.lang.String is made immutable. So why doesn't java maintain constant pools of other immutable types, such as Long, Integer, Char, Short ? Wouldn't that save memory too ?
I am aware of the fact that Integers are pooled for value range [-127, 127], though I do not understand the reason for choosing this range.
Here's a test code I wrote to test pooling of other immutable data types.
public class PoolTest {
public static void main(String... args) {
// Pooling of Integer [-127, 127]
Integer x = 127, y = 127;
System.out.println("Integer:" + (x == y)); // prints true
x = 129;
y = 129;
System.out.println("Integer:" + (x == y)); // prints false
// Apparent pooling of short [-127, 127]
Short i = 127, j = 127;
System.out.println("Short: " + (i == j)); // prints true
i = 128;
j = 128;
System.out.println("Short: " + (i == j)); // prints false
// No pooling of long values
Long k = 10L, l = 10L;
System.out.println("Long: " + (i == j)); // prints false
k = 128L;
l = 128L;
System.out.println("Long: " + (i == j)); // prints false
}
}

The purpose of a constant pool is to reduce the memory overhead required by keeping multiple copies of constants. In the case of Strings, the JVM is inherently required to keep some object around for each individually distinguishable constant, and the Java spec basically says that the JVM should deduplicate String objects when class loading. The ability to manually place Strings in the pool via intern is inexpensive and allows programmers to identify particular values (such as properties) that are going to be around for the life of the program and tell the JVM to put them out of the way of normal garbage collection.
Pooling numeric constants, on the other hand, doesn't make a lot of sense, for a few reasons:
Most particular numbers aren't ever used in a given program's code.
When numbers are used in code, embedding them in the code as immediate opcode values is less expensive in terms of memory than trying to pool them. Note that even the empty String carries around a char[], an int for its length, and another for its hashCode. For a number, by contrast, a maximum of eight immediate bytes is required.
As of recent Java versions, Byte, Short, and Integer objects from -128 to 127 (0 to 127 for Character) are precached for performance reasons, not to save memory. This range was presumably chosen because this is the ranged of a signed byte, and it will cover a large number of common uses, while it would be impractical to try to precache a very large number of values.
As a note, keep in mind that the rules about interning were made long before the introduction of autoboxing and generic types in Java 5, which significantly expanded how much the wrapper classes were casually used. This increase in use led Sun to add those common values to a constant pool.

Well, because String objects are immutable, it's safe for multiple references to "share" the same String object.
public class ImmutableStrings
{
public static void main(String[] args)
{
String one = "str1";
String two = "str1";
System.out.println(one.equals(two));
System.out.println(one == two);
}
}
// Output
true
true
In such a case, there is really no need to make two instances of an identical String object. If a String object could be changed, as a StringBuffer can be changed, we would be forced to create two separate objects. But, as we know that String objects cannot change, we can safely share a String object among the two String references, one and two. This is done through the String literal pool.
You can go through this link to know in details.

java == for Integer [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Inconsistent behavior on java's ==
Integer wrapper objects share the same instances only within the value 127?
I have found the following == behaviour for Integer objects and I fail to understand it. (I am well aware that one should use equals for such comparisons, but I am studying for OCPJP...)
On short, == works as expected for 1000, but not for 10.
The former fragment of code is:
Integer i1 = 1000;
Integer i2 = 1000;
if(i1 != i2) System.out.println("different objects");
if(i1.equals(i2)) System.out.println("meaningfully equal");
and it behaves as one would expect:
different objects
meaningfully equal
The latter though:
Integer i3 = 10;
Integer i4 = 10;
if(i3 == i4) System.out.println("same object");
if(i3.equals(i4)) System.out.println("meaningfully equal");
has the following output:
same object
meaningfully equal
Can someone please explain why this is happening?

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

See Integer wrapper objects share the same instances only within the value 127?. The Integer class keeps shared static instances around for common values.

Java Wrapper equality test

public class WrapperTest {
public static void main(String[] args) {
Integer i = 100;
Integer j = 100;
if(i == j)
System.out.println("same");
else
System.out.println("not same");
}
}
The above code gives the output of same when run, however if we change the value of i and j to 1000 the output changes to not same. As I'm preparing for SCJP, need to get the concept behind this clear. Can someone explain this behavior.Thanks.

In Java, Integers between -128 and 127 (inclusive) are generally represented by the same Integer object instance. This is handled by the use of a inner class called IntegerCache (contained inside the Integer class, and used e.g. when Integer.valueOf() is called, or during autoboxing):
private static class IntegerCache {
private IntegerCache(){}
static final Integer cache[] = new Integer[-(-128) + 127 + 1];
static {
for(int i = 0; i < cache.length; i++)
cache[i] = new Integer(i - 128);
}
}
See also: http://www.owasp.org/index.php/Java_gotchas

Basically Integers between -127 and 127 are 'cached' in such a way that when you use those numbers you always refer to the same number in memory, which is why your == works.
Any Integer outside of that range are not cached, thus the references are not the same.

#tunaranch is correct. It is also the same issue as in this Python question. The gist is that Java keeps an object around for the integers from -128 to 127 (Python does -5 to 256) and returns the same object every time you ask for one. If you ask for an Integer outside of this fixed range, it'll give you a new object every time.
(Recall that == returns whether two objects are actually the same, while equals compares their contents.)
Edit: Here's the relevant paragraph from Section 5.1.7 of the Java Language Specification:
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.
Note that this also describes what happens with other types.

It's to do with equality and autoboxing: http://web.archive.org/web/20090220142800/http://davidflanagan.com/2004/02/equality-and-autoboxing.html

Your code doesn't compile. This is what I get:
Exception in thread "main" java.lang.Error: Unresolved compilation problems:
Type mismatch: cannot convert from int to Integer
Type mismatch: cannot convert from int to Integer
at WrapperTest.main(WrapperTest.java:5)
Variables i and j are instances of Integer object. Don't compare instances of object using "==" operator, use "equals" method instead.
Greetings