Develop Reference

Why if condition return false on equal same values [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() {}
}

Regarding equals() and == in a generic class [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() {}
}

Comparation doesnt work for certain values [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() {}
}

Boundary of 127 for integers in Java [duplicate]

class D {
public static void main(String args[]) {
Integer b2=128;
Integer b3=128;
System.out.println(b2==b3);
}
}
Output:
false
class D {
public static void main(String args[]) {
Integer b2=127;
Integer b3=127;
System.out.println(b2==b3);
}
}
Output:
true
Note: Numbers between -128 and 127 are true.

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.

Autoboxing caches -128 to 127. This is specified in the JLS (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.
A simple rule to remember when dealing with objects is - use .equals if you want to check if the two objects are "equal", use == when you want to see if they point to the same instance.

Using primitive data types, ints, would produce true in both cases, the expected output.
However, since you're using Integer objects the == operator has a different meaning.
In the context of objects, == checks to see if the variables refer to the same object reference.
To compare the value of the objects you should use the equals() method
E.g.
b2.equals(b1)
which will indicate whether b2 is less than b1, greater than, or equal to (check the API for details)

It is memory optimization in Java related.
To save on memory, Java 'reuses' all the wrapper objects whose values
fall in the following ranges:
All Boolean values (true and false)
All Byte values
All Character values from \u0000 to \u007f (i.e. 0 to 127 in decimal)
All Short and Integer values from -128 to 127.

Have a look at the Integer.java, if the value is between -128 and 127, it will use the cached pool, so (Integer) 1 == (Integer) 1 while (Integer) 222 != (Integer) 222
/**
* 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) {
assert IntegerCache.high >= 127;
if (i >= IntegerCache.low && i <= IntegerCache.high)
return IntegerCache.cache[i + (-IntegerCache.low)];
return new Integer(i);
}

Other answers describe why the observed effects can be observed, but that's really beside the point for programmers (interesting, certainly, but something you should forget all about when writing actual code).
To compare Integer objects for equality, use the equals method.
Do not attempt to compare Integer objects for equality by using the identity operator, ==.
It may happen that some equal values are identical objects, but this is not something that should generally be relied on.

if the value is between -128 and 127, it will use the cached pool and this is true only when auto-boxing.
So you will have below:
public static void main(String[] args) {
Integer a = new Integer(100);
Integer b = new Integer(100);
System.out.println(a == b); // false. == compare two instances, they are difference
System.out.println(a.equals(b)); // true. equals compares the value
Integer a2 = 100;
Integer b2 = 100;
System.out.println(a2 == b2); // true. auto-boxing uses cached pool between -128/127
System.out.println(a2.equals(b2)); // true. equals compares the value
Integer a3 = 129;
Integer b3 = 129;
System.out.println(a3 == b3); // false. not using cached pool
System.out.println(a3.equals(b3)); // true. equals compares the value
}
}

I wrote the following as this problem isn't just specific to Integer. My conclusion is that more often than not if you use the API incorrectly, you sill see incorrect behavior. Use it correctly and you should see the correct behavior:
public static void main (String[] args) {
Byte b1=127;
Byte b2=127;
Short s1=127; //incorrect should use Byte
Short s2=127; //incorrect should use Byte
Short s3=128;
Short s4=128;
Integer i1=127; //incorrect should use Byte
Integer i2=127; //incorrect should use Byte
Integer i3=128;
Integer i4=128;
Integer i5=32767; //incorrect should use Short
Integer i6=32767; //incorrect should use Short
Long l1=127L; //incorrect should use Byte
Long l2=127L; //incorrect should use Byte
Long l3=13267L; //incorrect should use Short
Long l4=32767L; //incorrect should use Short
Long l5=2147483647L; //incorrect should use Integer
Long l6=2147483647L; //incorrect should use Integer
Long l7=2147483648L;
Long l8=2147483648L;
System.out.print(b1==b2); //true (incorrect) Used API correctly
System.out.print(s1==s2); //true (incorrect) Used API incorrectly
System.out.print(i1==i2); //true (incorrect) Used API incorrectly
System.out.print(l1==l2); //true (incorrect) Used API incorrectly
System.out.print(s3==s4); //false (correct) Used API correctly
System.out.print(i3==i4); //false (correct) Used API correctly
System.out.print(i5==i6); //false (correct) Used API correctly
System.out.print(l3==l4); //false (correct) Used API correctly
System.out.print(l7==l8); //false (correct) Used API correctly
System.out.print(l5==l6); //false (correct) Used API incorrectly
}

Why is the output of following code different in the two print statements? [duplicate]

class D {
public static void main(String args[]) {
Integer b2=128;
Integer b3=128;
System.out.println(b2==b3);
}
}
Output:
false
class D {
public static void main(String args[]) {
Integer b2=127;
Integer b3=127;
System.out.println(b2==b3);
}
}
Output:
true
Note: Numbers between -128 and 127 are true.

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.

Autoboxing caches -128 to 127. This is specified in the JLS (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.
A simple rule to remember when dealing with objects is - use .equals if you want to check if the two objects are "equal", use == when you want to see if they point to the same instance.

Using primitive data types, ints, would produce true in both cases, the expected output.
However, since you're using Integer objects the == operator has a different meaning.
In the context of objects, == checks to see if the variables refer to the same object reference.
To compare the value of the objects you should use the equals() method
E.g.
b2.equals(b1)
which will indicate whether b2 is less than b1, greater than, or equal to (check the API for details)

It is memory optimization in Java related.
To save on memory, Java 'reuses' all the wrapper objects whose values
fall in the following ranges:
All Boolean values (true and false)
All Byte values
All Character values from \u0000 to \u007f (i.e. 0 to 127 in decimal)
All Short and Integer values from -128 to 127.

Have a look at the Integer.java, if the value is between -128 and 127, it will use the cached pool, so (Integer) 1 == (Integer) 1 while (Integer) 222 != (Integer) 222
/**
* 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) {
assert IntegerCache.high >= 127;
if (i >= IntegerCache.low && i <= IntegerCache.high)
return IntegerCache.cache[i + (-IntegerCache.low)];
return new Integer(i);
}

Other answers describe why the observed effects can be observed, but that's really beside the point for programmers (interesting, certainly, but something you should forget all about when writing actual code).
To compare Integer objects for equality, use the equals method.
Do not attempt to compare Integer objects for equality by using the identity operator, ==.
It may happen that some equal values are identical objects, but this is not something that should generally be relied on.

if the value is between -128 and 127, it will use the cached pool and this is true only when auto-boxing.
So you will have below:
public static void main(String[] args) {
Integer a = new Integer(100);
Integer b = new Integer(100);
System.out.println(a == b); // false. == compare two instances, they are difference
System.out.println(a.equals(b)); // true. equals compares the value
Integer a2 = 100;
Integer b2 = 100;
System.out.println(a2 == b2); // true. auto-boxing uses cached pool between -128/127
System.out.println(a2.equals(b2)); // true. equals compares the value
Integer a3 = 129;
Integer b3 = 129;
System.out.println(a3 == b3); // false. not using cached pool
System.out.println(a3.equals(b3)); // true. equals compares the value
}
}

I wrote the following as this problem isn't just specific to Integer. My conclusion is that more often than not if you use the API incorrectly, you sill see incorrect behavior. Use it correctly and you should see the correct behavior:
public static void main (String[] args) {
Byte b1=127;
Byte b2=127;
Short s1=127; //incorrect should use Byte
Short s2=127; //incorrect should use Byte
Short s3=128;
Short s4=128;
Integer i1=127; //incorrect should use Byte
Integer i2=127; //incorrect should use Byte
Integer i3=128;
Integer i4=128;
Integer i5=32767; //incorrect should use Short
Integer i6=32767; //incorrect should use Short
Long l1=127L; //incorrect should use Byte
Long l2=127L; //incorrect should use Byte
Long l3=13267L; //incorrect should use Short
Long l4=32767L; //incorrect should use Short
Long l5=2147483647L; //incorrect should use Integer
Long l6=2147483647L; //incorrect should use Integer
Long l7=2147483648L;
Long l8=2147483648L;
System.out.print(b1==b2); //true (incorrect) Used API correctly
System.out.print(s1==s2); //true (incorrect) Used API incorrectly
System.out.print(i1==i2); //true (incorrect) Used API incorrectly
System.out.print(l1==l2); //true (incorrect) Used API incorrectly
System.out.print(s3==s4); //false (correct) Used API correctly
System.out.print(i3==i4); //false (correct) Used API correctly
System.out.print(i5==i6); //false (correct) Used API correctly
System.out.print(l3==l4); //false (correct) Used API correctly
System.out.print(l7==l8); //false (correct) Used API correctly
System.out.print(l5==l6); //false (correct) Used API incorrectly
}