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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() {}
}

Clarification regarding Integer comparison? [duplicate]

This question already has answers here:
Boxed Primitives and Equivalence
(4 answers)
Closed 9 years ago.
class Demo{
public static void main(String[] args) {
Integer i = Integer.valueOf(127);
Integer j = Integer.valueOf(127);
System.out.println(i==j);
Integer k = Integer.valueOf(128);
Integer l = Integer.valueOf(128);
System.out.println(k==l);
}
}
The first print statement prints true whereas the second one prints false.Why?
Please explain in detail.

It is because Integer caching.
From java language specification 5.1.7
If the value p being boxed is true, false, a byte, or a char in the range
\u0000 to \u007f, or an int or short number between -128 and 127 (inclusive),
then let r1 and r2 be the results of any two boxing conversions of p.
It is always the case that r1 == r2.
Ideally, boxing a given primitive value p, would always yield an identical reference.
Integer i = Integer.valueOf(127);
Integer j = Integer.valueOf(127);
Both i and j point to same object. As the value is less than 127.
Integer k = Integer.valueOf(128);
Integer l = Integer.valueOf(128);
Both k & l point to different objects. As the value is greater than 127.
As, you are checking the object references using == operator, you are getting different results.
Update
You can use equals() method to get the same result
System.out.println(i.equals(j));//equals() compares the values of objects not references
System.out.println(k.equals(l));//equals() compares the values of objects not references
Output is
true
true
== operator checks the actual object references.
equals() checks the values(contents) of objects.
Answer to comment
You have,
Integer i = Integer.valueOf(127);
Here new object is created & reference is assigned to i
Integer j = Integer.valueOf(127); //will not create new object as it already exists
Due to integer caching (number between -128 to 127) previously created object reference is assigned to j, then i and j point to same objects.
Now consider,
Integer p = Integer.valueOf(127); //create new object
Integer q = Integer.valueOf(126); //this also creates new object as it does not exists
Obviously both checks using == operator and equals() method will result false. As both are different references and have different vales.

i==j
is true for values between -128 and 127 due to integer caching.
From language spec
If the value p being boxed is true, false, a byte, or a char in the range \u0000 to \u007f, or an int or short number between -128 and 127 (inclusive), then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.
Integer i = Integer.valueOf(127); // new object
Integer j = Integer.valueOf(127); //cached object reference
Integer k = Integer.valueOf(128); // new object
Integer l = Integer.valueOf(128); // new object
So i and j are pointing to same reference because of value 127.
Where as k and l pointing to difference references, because their value >127
There is a reason mentioned in docs for this behaviour:
The behavior will be the desired one, without imposing an undue performance penalty, especially on small devices. Less memory-limited implementations might

valueOf returns an Integer Object. Integer is a wrapper class for int. For your case,
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.
You fix it by following way if you want same result for your both cases:
Make the types int
Cast the types to int or
Use .equals()

Why the First if case prints Equal and Second Not Equal? [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Weird Java Boxing
public class example {
public static void main(String[] args) {
Integer a=127,b=127;
if(a==b)
System.out.println("Equal");
else
System.out.println("Not Equal");
Integer c=128,d=128;
if(c==d)
System.out.println("Equal");
else
System.out.println("Not Equal");
}
}
Output:
Equal
Not Equal

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.
Hence when you tried to compare 127 with 127 there was only one object
made and it worked right but when you tried with 128 it came out of
the range and it created two objects so you can't compare them using
== operator.
For this purpose use .equals()(Comparing Object reference) method.Please refer
this for more details .
Integer c=128,d=128;
if(c.equals(d))
System.out.println("Equal");
else
System.out.println("Not Equal");

This happens because Integer numbers within range [-128, 127] are stored in a heap. Thus, as you compare object references and not object values and both Integer a = 127 and Integer b = 127 point to the same object in a heap your expression is evaluated to true.
When Integer values are out of that range, you get two different objects, so reference comparison returns false

From the source code of Integer.java
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);
}
}
/**
* Returns a <tt>Integer</tt> instance representing the specified
* <tt>int</tt> value.
* If a new <tt>Integer</tt> 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.
*
* #param i an <code>int</code> value.
* #return a <tt>Integer</tt> instance representing <tt>i</tt>.
* #since 1.5
*/
public static Integer valueOf(int i) {
final int offset = 128;
if (i >= -128 && i <= 127) { // must cache
return IntegerCache.cache[i + offset];
}
return new Integer(i);
}

Boxed Primitives and Equivalence

So I was asked this question today.
Integer a = 3;
Integer b = 2;
Integer c = 5;
Integer d = a + b;
System.out.println(c == d);
What will this program print out? It returns true. I answered it will always print out false because of how I understood auto (and auto un) boxing. I was under the impression that assigning Integer a = 3 will create a new Integer(3) so that an == will evaluate the reference rather then the primitive value.
Can anyone explain this?

Boxed values between -128 to 127 are cached. Boxing uses Integer.valueOf method, which uses the cache. Values outside the range are not cached and always created as a new instance. Since your values fall into the cached range, values are equal using == operator.
Quote from 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.
http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.7

This is what is really happening:
Integer c = Integer.valueOf(5);
Integer d = Integer.valueOf(a.intValue() + b.intValue());
Java maintains a cache of Integer objects between -128 and 127. Compare with the following:
Integer a = 300;
Integer b = 200;
Integer c = 500;
Integer d = a + b;
System.out.println(c == d);
Which should print false.

It's because some of the (auto-boxed) Integers are cached, so you're actually comparing the same reference -- this post has more detailed examples and an explanation.

Caching happens outside of autoboxing too, consider this:
Integer a = 1;
Integer b = new Integer(1);
Integer c = Integer.valueOf(1);
System.out.println(a == b);
System.out.println(b == c);
System.out.println(c == a);
this will print:
false
false
true
Generally you want to stay away from == when comparing Objects

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