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Integer caching in Java with new operator


Weird Integer boxing in JavaWhy aren't Integers cached in Java?Is Java “pass-by-reference” or “pass-by-value”?How do I efficiently iterate over each entry in a Java Map?What is the difference between public, protected, package-private and private in Java?How do I read / convert an InputStream into a String in Java?When to use LinkedList over ArrayList in Java?How do I generate random integers within a specific range in Java?Efficiency of Java “Double Brace Initialization”?How do I convert a String to an int in Java?Creating a memory leak with JavaWhy don't Java's +=, -=, *=, /= compound assignment operators require casting?






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18















In the below class I have tried to compare the wrapper class with the primitive but the results are different.



I have checked the following links links:




The more interesting question is why new Object(); should be required to create a unique instance every time? i. e. why is new Object(); not allowed to cache? The answer is the wait(...) and notify(...) calls. Caching new Object()s would incorrectly cause threads to synchronize with each other when they shouldn't.




If there is a new object then how are a and c equal?



If b is equal to c and c is equal to a, then a should be equal to b. But in following case I got a != c.



Please explain.



class WrapperCompare 

public static void main (String args[])
Integer a = new Integer(10);
Integer b = 10;
int c=10;
System.out.println(b==c); //true
System.out.println(a==b); //false
System.out.println(a==c); //true




Update:
By referring to this link Integer caching.




Basically, the Integer class keeps a cache of Integer instances in the range of -128 to 127, and all autoboxing, literals and uses of Integer.valueOf() will return instances from that cache for the range it covers.




So in this case all statements should be true.










share|improve this question



















  • 2





    Note that the constructor is deprecated and should not be used. Instead use valueOf.

    – Zabuza
    May 7 at 6:52






  • 5





    All but a==b here aren't comparing objects.

    – Andy Turner
    May 7 at 6:53






  • 5





    @Zabuza that's incorrect. new always returns a new instance. You only get the cached instances via valueOf (implicitly or explicitly).

    – Andy Turner
    May 7 at 6:54







  • 3





    The statement Integer b = 10; is equivalent to Integer b = Integer.valueOf(10); (and not Integer b = new Integer(10);)

    – Maurice Perry
    May 7 at 6:58







  • 1





    Good question, good answer, good comments. Good job everyone! :)

    – JollyJoker
    May 7 at 9:00

















18















In the below class I have tried to compare the wrapper class with the primitive but the results are different.



I have checked the following links links:




The more interesting question is why new Object(); should be required to create a unique instance every time? i. e. why is new Object(); not allowed to cache? The answer is the wait(...) and notify(...) calls. Caching new Object()s would incorrectly cause threads to synchronize with each other when they shouldn't.




If there is a new object then how are a and c equal?



If b is equal to c and c is equal to a, then a should be equal to b. But in following case I got a != c.



Please explain.



class WrapperCompare 

public static void main (String args[])
Integer a = new Integer(10);
Integer b = 10;
int c=10;
System.out.println(b==c); //true
System.out.println(a==b); //false
System.out.println(a==c); //true




Update:
By referring to this link Integer caching.




Basically, the Integer class keeps a cache of Integer instances in the range of -128 to 127, and all autoboxing, literals and uses of Integer.valueOf() will return instances from that cache for the range it covers.




So in this case all statements should be true.










share|improve this question



















  • 2





    Note that the constructor is deprecated and should not be used. Instead use valueOf.

    – Zabuza
    May 7 at 6:52






  • 5





    All but a==b here aren't comparing objects.

    – Andy Turner
    May 7 at 6:53






  • 5





    @Zabuza that's incorrect. new always returns a new instance. You only get the cached instances via valueOf (implicitly or explicitly).

    – Andy Turner
    May 7 at 6:54







  • 3





    The statement Integer b = 10; is equivalent to Integer b = Integer.valueOf(10); (and not Integer b = new Integer(10);)

    – Maurice Perry
    May 7 at 6:58







  • 1





    Good question, good answer, good comments. Good job everyone! :)

    – JollyJoker
    May 7 at 9:00













18












18








18


2






In the below class I have tried to compare the wrapper class with the primitive but the results are different.



I have checked the following links links:




The more interesting question is why new Object(); should be required to create a unique instance every time? i. e. why is new Object(); not allowed to cache? The answer is the wait(...) and notify(...) calls. Caching new Object()s would incorrectly cause threads to synchronize with each other when they shouldn't.




If there is a new object then how are a and c equal?



If b is equal to c and c is equal to a, then a should be equal to b. But in following case I got a != c.



Please explain.



class WrapperCompare 

public static void main (String args[])
Integer a = new Integer(10);
Integer b = 10;
int c=10;
System.out.println(b==c); //true
System.out.println(a==b); //false
System.out.println(a==c); //true




Update:
By referring to this link Integer caching.




Basically, the Integer class keeps a cache of Integer instances in the range of -128 to 127, and all autoboxing, literals and uses of Integer.valueOf() will return instances from that cache for the range it covers.




So in this case all statements should be true.










share|improve this question
















In the below class I have tried to compare the wrapper class with the primitive but the results are different.



I have checked the following links links:




The more interesting question is why new Object(); should be required to create a unique instance every time? i. e. why is new Object(); not allowed to cache? The answer is the wait(...) and notify(...) calls. Caching new Object()s would incorrectly cause threads to synchronize with each other when they shouldn't.




If there is a new object then how are a and c equal?



If b is equal to c and c is equal to a, then a should be equal to b. But in following case I got a != c.



Please explain.



class WrapperCompare 

public static void main (String args[])
Integer a = new Integer(10);
Integer b = 10;
int c=10;
System.out.println(b==c); //true
System.out.println(a==b); //false
System.out.println(a==c); //true




Update:
By referring to this link Integer caching.




Basically, the Integer class keeps a cache of Integer instances in the range of -128 to 127, and all autoboxing, literals and uses of Integer.valueOf() will return instances from that cache for the range it covers.




So in this case all statements should be true.







java instance equals






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited May 7 at 14:43









Boann

37.8k1291123




37.8k1291123










asked May 7 at 6:47









User12345User12345

14412




14412







  • 2





    Note that the constructor is deprecated and should not be used. Instead use valueOf.

    – Zabuza
    May 7 at 6:52






  • 5





    All but a==b here aren't comparing objects.

    – Andy Turner
    May 7 at 6:53






  • 5





    @Zabuza that's incorrect. new always returns a new instance. You only get the cached instances via valueOf (implicitly or explicitly).

    – Andy Turner
    May 7 at 6:54







  • 3





    The statement Integer b = 10; is equivalent to Integer b = Integer.valueOf(10); (and not Integer b = new Integer(10);)

    – Maurice Perry
    May 7 at 6:58







  • 1





    Good question, good answer, good comments. Good job everyone! :)

    – JollyJoker
    May 7 at 9:00












  • 2





    Note that the constructor is deprecated and should not be used. Instead use valueOf.

    – Zabuza
    May 7 at 6:52






  • 5





    All but a==b here aren't comparing objects.

    – Andy Turner
    May 7 at 6:53






  • 5





    @Zabuza that's incorrect. new always returns a new instance. You only get the cached instances via valueOf (implicitly or explicitly).

    – Andy Turner
    May 7 at 6:54







  • 3





    The statement Integer b = 10; is equivalent to Integer b = Integer.valueOf(10); (and not Integer b = new Integer(10);)

    – Maurice Perry
    May 7 at 6:58







  • 1





    Good question, good answer, good comments. Good job everyone! :)

    – JollyJoker
    May 7 at 9:00







2




2





Note that the constructor is deprecated and should not be used. Instead use valueOf.

– Zabuza
May 7 at 6:52





Note that the constructor is deprecated and should not be used. Instead use valueOf.

– Zabuza
May 7 at 6:52




5




5





All but a==b here aren't comparing objects.

– Andy Turner
May 7 at 6:53





All but a==b here aren't comparing objects.

– Andy Turner
May 7 at 6:53




5




5





@Zabuza that's incorrect. new always returns a new instance. You only get the cached instances via valueOf (implicitly or explicitly).

– Andy Turner
May 7 at 6:54






@Zabuza that's incorrect. new always returns a new instance. You only get the cached instances via valueOf (implicitly or explicitly).

– Andy Turner
May 7 at 6:54





3




3





The statement Integer b = 10; is equivalent to Integer b = Integer.valueOf(10); (and not Integer b = new Integer(10);)

– Maurice Perry
May 7 at 6:58






The statement Integer b = 10; is equivalent to Integer b = Integer.valueOf(10); (and not Integer b = new Integer(10);)

– Maurice Perry
May 7 at 6:58





1




1





Good question, good answer, good comments. Good job everyone! :)

– JollyJoker
May 7 at 9:00





Good question, good answer, good comments. Good job everyone! :)

– JollyJoker
May 7 at 9:00












1 Answer
1






active

oldest

votes


















25














Explanation



When you compare Integer vs int with ==, it needs to convert the Integer to an int. This is called unboxing.



See JLS§5.1.8:




If r is a reference of type Integer, then unboxing conversion converts r into r.intValue()




At that point, you are comparing int vs int. And primitives have no notion of instances, they all refer to the same value. As such, the result is true.



So the actual code you have is



a.intValue() == c


leading to a comparison of 10 == 10, both int values, no Integer instances anymore.



You can see that new Integer(...) indeed creates new instances, when you compare Integer vs Integer. You did that in a == b.




Note



The constructor new Integer(...) is deprecated. You should instead use Integer#valueOf, it is potentially faster and also uses an internal cache. From the documentation:




Returns an Integer instance representing the specified int value. If a new Integer instance is not required, this method should generally be used in preference to the constructor 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.




The caching is important to note here, since it yields to == being true again (for cached values):



Integer first = Integer.valueOf(10);
Integer second = Integer.valueOf(10);
System.out.println(first == second); // true


The caching is guaranteed for values between -128 and +127, but may also be used for others.



Also note that your b actually comes out of the cache, since



Integer b = 10;
// same as
Integer b = Integer.valueOf(10);
// and not
Integer b = new Integer(10);


So boxing goes through Integers cache (see JLS§5.1.7).






share|improve this answer




















  • 2





    “it is much faster” is a too strong claim. As the documentation says, “this method is likely to yield significantly better space and time performance”, in other words, it enables potential optimizations, whether they exist, is implementation dependent.

    – Holger
    May 7 at 10:36











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









25














Explanation



When you compare Integer vs int with ==, it needs to convert the Integer to an int. This is called unboxing.



See JLS§5.1.8:




If r is a reference of type Integer, then unboxing conversion converts r into r.intValue()




At that point, you are comparing int vs int. And primitives have no notion of instances, they all refer to the same value. As such, the result is true.



So the actual code you have is



a.intValue() == c


leading to a comparison of 10 == 10, both int values, no Integer instances anymore.



You can see that new Integer(...) indeed creates new instances, when you compare Integer vs Integer. You did that in a == b.




Note



The constructor new Integer(...) is deprecated. You should instead use Integer#valueOf, it is potentially faster and also uses an internal cache. From the documentation:




Returns an Integer instance representing the specified int value. If a new Integer instance is not required, this method should generally be used in preference to the constructor 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.




The caching is important to note here, since it yields to == being true again (for cached values):



Integer first = Integer.valueOf(10);
Integer second = Integer.valueOf(10);
System.out.println(first == second); // true


The caching is guaranteed for values between -128 and +127, but may also be used for others.



Also note that your b actually comes out of the cache, since



Integer b = 10;
// same as
Integer b = Integer.valueOf(10);
// and not
Integer b = new Integer(10);


So boxing goes through Integers cache (see JLS§5.1.7).






share|improve this answer




















  • 2





    “it is much faster” is a too strong claim. As the documentation says, “this method is likely to yield significantly better space and time performance”, in other words, it enables potential optimizations, whether they exist, is implementation dependent.

    – Holger
    May 7 at 10:36















25














Explanation



When you compare Integer vs int with ==, it needs to convert the Integer to an int. This is called unboxing.



See JLS§5.1.8:




If r is a reference of type Integer, then unboxing conversion converts r into r.intValue()




At that point, you are comparing int vs int. And primitives have no notion of instances, they all refer to the same value. As such, the result is true.



So the actual code you have is



a.intValue() == c


leading to a comparison of 10 == 10, both int values, no Integer instances anymore.



You can see that new Integer(...) indeed creates new instances, when you compare Integer vs Integer. You did that in a == b.




Note



The constructor new Integer(...) is deprecated. You should instead use Integer#valueOf, it is potentially faster and also uses an internal cache. From the documentation:




Returns an Integer instance representing the specified int value. If a new Integer instance is not required, this method should generally be used in preference to the constructor 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.




The caching is important to note here, since it yields to == being true again (for cached values):



Integer first = Integer.valueOf(10);
Integer second = Integer.valueOf(10);
System.out.println(first == second); // true


The caching is guaranteed for values between -128 and +127, but may also be used for others.



Also note that your b actually comes out of the cache, since



Integer b = 10;
// same as
Integer b = Integer.valueOf(10);
// and not
Integer b = new Integer(10);


So boxing goes through Integers cache (see JLS§5.1.7).






share|improve this answer




















  • 2





    “it is much faster” is a too strong claim. As the documentation says, “this method is likely to yield significantly better space and time performance”, in other words, it enables potential optimizations, whether they exist, is implementation dependent.

    – Holger
    May 7 at 10:36













25












25








25







Explanation



When you compare Integer vs int with ==, it needs to convert the Integer to an int. This is called unboxing.



See JLS§5.1.8:




If r is a reference of type Integer, then unboxing conversion converts r into r.intValue()




At that point, you are comparing int vs int. And primitives have no notion of instances, they all refer to the same value. As such, the result is true.



So the actual code you have is



a.intValue() == c


leading to a comparison of 10 == 10, both int values, no Integer instances anymore.



You can see that new Integer(...) indeed creates new instances, when you compare Integer vs Integer. You did that in a == b.




Note



The constructor new Integer(...) is deprecated. You should instead use Integer#valueOf, it is potentially faster and also uses an internal cache. From the documentation:




Returns an Integer instance representing the specified int value. If a new Integer instance is not required, this method should generally be used in preference to the constructor 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.




The caching is important to note here, since it yields to == being true again (for cached values):



Integer first = Integer.valueOf(10);
Integer second = Integer.valueOf(10);
System.out.println(first == second); // true


The caching is guaranteed for values between -128 and +127, but may also be used for others.



Also note that your b actually comes out of the cache, since



Integer b = 10;
// same as
Integer b = Integer.valueOf(10);
// and not
Integer b = new Integer(10);


So boxing goes through Integers cache (see JLS§5.1.7).






share|improve this answer















Explanation



When you compare Integer vs int with ==, it needs to convert the Integer to an int. This is called unboxing.



See JLS§5.1.8:




If r is a reference of type Integer, then unboxing conversion converts r into r.intValue()




At that point, you are comparing int vs int. And primitives have no notion of instances, they all refer to the same value. As such, the result is true.



So the actual code you have is



a.intValue() == c


leading to a comparison of 10 == 10, both int values, no Integer instances anymore.



You can see that new Integer(...) indeed creates new instances, when you compare Integer vs Integer. You did that in a == b.




Note



The constructor new Integer(...) is deprecated. You should instead use Integer#valueOf, it is potentially faster and also uses an internal cache. From the documentation:




Returns an Integer instance representing the specified int value. If a new Integer instance is not required, this method should generally be used in preference to the constructor 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.




The caching is important to note here, since it yields to == being true again (for cached values):



Integer first = Integer.valueOf(10);
Integer second = Integer.valueOf(10);
System.out.println(first == second); // true


The caching is guaranteed for values between -128 and +127, but may also be used for others.



Also note that your b actually comes out of the cache, since



Integer b = 10;
// same as
Integer b = Integer.valueOf(10);
// and not
Integer b = new Integer(10);


So boxing goes through Integers cache (see JLS§5.1.7).







share|improve this answer














share|improve this answer



share|improve this answer








edited May 7 at 14:50

























answered May 7 at 6:57









ZabuzaZabuza

12.6k62945




12.6k62945







  • 2





    “it is much faster” is a too strong claim. As the documentation says, “this method is likely to yield significantly better space and time performance”, in other words, it enables potential optimizations, whether they exist, is implementation dependent.

    – Holger
    May 7 at 10:36












  • 2





    “it is much faster” is a too strong claim. As the documentation says, “this method is likely to yield significantly better space and time performance”, in other words, it enables potential optimizations, whether they exist, is implementation dependent.

    – Holger
    May 7 at 10:36







2




2





“it is much faster” is a too strong claim. As the documentation says, “this method is likely to yield significantly better space and time performance”, in other words, it enables potential optimizations, whether they exist, is implementation dependent.

– Holger
May 7 at 10:36





“it is much faster” is a too strong claim. As the documentation says, “this method is likely to yield significantly better space and time performance”, in other words, it enables potential optimizations, whether they exist, is implementation dependent.

– Holger
May 7 at 10:36



















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