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How is it possible to add a double into an ArrayList of Integer? (Java)

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16

2

I try to understand how is it possible to have a Double value into an ArrayList of Integer. The numList is an ArrayList of Integer, and the value from it is a Double.

This is the code:

package bounded.wildcards;

import java.util.ArrayList;
import java.util.List;

public class GenericsDemo 

 public static void main(String[] args) 
 // Invariance Workaround
 List<Integer> numList = new ArrayList<>();
 GenericsDemo.invarianceWorkaround(numList);
 System.out.println(numList);
 

 static <T extends Number> void invarianceWorkaround(List<T> list) 

 T element = (T) new Double(23.3); 
 list.add(element);
 

This will compile and run without an error.

java generics arraylist

share|improve this question

edited Mar 28 at 13:18

gaby

asked Mar 28 at 11:45

gabygaby

595112

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Possible duplicate of type erasure in implementation of ArrayList in Java
  
  – vaxquis
  Mar 28 at 13:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  also, stackoverflow.com/questions/339699/… and many others... I think that someone can find an exact duplicate, but they are are extremely closely related.
  
  – vaxquis
  Mar 28 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for the links, I know about type erasure and add cast, but I cannot understand how is it possible to have an ArrayList of Integer with this Double value in it. And there is no exception when I run it.
  
  – gaby
  Mar 28 at 18:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @gabv you're contradicting yourself. If you knew about type erasure, you'd understand why you can add anything to ArrayList<Double> at runtime... because the answer is because of type erasure. As such, it follows you must either not know about type erasure mechanism in Java or not understand it, which both mean you should learn more about it.
  
  – vaxquis
  Mar 28 at 21:20
  
  

  
  
  
  

add a comment | 

16

2

I try to understand how is it possible to have a Double value into an ArrayList of Integer. The numList is an ArrayList of Integer, and the value from it is a Double.

This is the code:

package bounded.wildcards;

import java.util.ArrayList;
import java.util.List;

public class GenericsDemo 

 public static void main(String[] args) 
 // Invariance Workaround
 List<Integer> numList = new ArrayList<>();
 GenericsDemo.invarianceWorkaround(numList);
 System.out.println(numList);
 

 static <T extends Number> void invarianceWorkaround(List<T> list) 

 T element = (T) new Double(23.3); 
 list.add(element);
 

This will compile and run without an error.

java generics arraylist

share|improve this question

edited Mar 28 at 13:18

gaby

asked Mar 28 at 11:45

gabygaby

595112

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Possible duplicate of type erasure in implementation of ArrayList in Java
  
  – vaxquis
  Mar 28 at 13:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  also, stackoverflow.com/questions/339699/… and many others... I think that someone can find an exact duplicate, but they are are extremely closely related.
  
  – vaxquis
  Mar 28 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for the links, I know about type erasure and add cast, but I cannot understand how is it possible to have an ArrayList of Integer with this Double value in it. And there is no exception when I run it.
  
  – gaby
  Mar 28 at 18:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @gabv you're contradicting yourself. If you knew about type erasure, you'd understand why you can add anything to ArrayList<Double> at runtime... because the answer is because of type erasure. As such, it follows you must either not know about type erasure mechanism in Java or not understand it, which both mean you should learn more about it.
  
  – vaxquis
  Mar 28 at 21:20
  
  

  
  
  
  

add a comment | 

I try to understand how is it possible to have a Double value into an ArrayList of Integer. The numList is an ArrayList of Integer, and the value from it is a Double.

This is the code:

package bounded.wildcards;

import java.util.ArrayList;
import java.util.List;

public class GenericsDemo 

 public static void main(String[] args) 
 // Invariance Workaround
 List<Integer> numList = new ArrayList<>();
 GenericsDemo.invarianceWorkaround(numList);
 System.out.println(numList);
 

 static <T extends Number> void invarianceWorkaround(List<T> list) 

 T element = (T) new Double(23.3); 
 list.add(element);
 

This will compile and run without an error.

java generics arraylist

share|improve this question

edited Mar 28 at 13:18

gaby

asked Mar 28 at 11:45

gabygaby

595112

package bounded.wildcards;

import java.util.ArrayList;
import java.util.List;

public class GenericsDemo 

 public static void main(String[] args) 
 // Invariance Workaround
 List<Integer> numList = new ArrayList<>();
 GenericsDemo.invarianceWorkaround(numList);
 System.out.println(numList);
 

 static <T extends Number> void invarianceWorkaround(List<T> list) 

 T element = (T) new Double(23.3); 
 list.add(element);
 

share|improve this question

edited Mar 28 at 13:18

gaby

asked Mar 28 at 11:45

gabygaby

595112

share|improve this question

edited Mar 28 at 13:18

gaby

asked Mar 28 at 11:45

gabygaby

595112

asked Mar 28 at 11:45

gabygaby

595112

asked Mar 28 at 11:45

gabygaby

595112

1 Answer
1

active

oldest

votes

21

This is because of type erasure used with Java generics - the type checks are only performed at compile time for generic types, and the type info for generics is then erased, effectively turning List<Integer> into List<Object>.

My IDE warns you of an "Unchecked cast from Double to T". But the compiler couldn't be sure that your code is wrong, so it does not emit an error, just a warning.

Then at runtime, the type check is no longer present due to type erasure, so the code will run without error unless you perform some operation that fails due to incompatible runtime type. System.out.println() is not such operation.

---

If you change the print code to

Integer num = numList.get(0);
System.out.println(num);

this will now involve runtime type check and will therefore fail:

java.lang.ClassCastException: java.lang.Double cannot be cast to java.lang.Integer

share|improve this answer

edited Mar 28 at 12:05

answered Mar 28 at 11:55

Jiri TousekJiri Tousek

10.7k52240

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Note that a ClassCastException is emitted when one tries to do this: Integer i = numList.get(0).
  
  – MC Emperor
  Mar 28 at 11:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MCEmperor Thanks, added. I couldn't find a simple example to force type incompatibility - integer seems to basically have no meaningful methods that wouldn't either be static or already in Number.
  
  – Jiri Tousek
  Mar 28 at 12:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @JiriTousek You may be interested in Java is Unsound. Basically all java compilers will add cast checks in places to avoid issues because they know that the type system is broken.
  
  – Giacomo Alzetta
  Mar 28 at 12:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  effectively turning into List<Number> not List<Object> because of upper bound is Number. if its List<Object> then you can store String but you can't
  
  – Akash Shah
  Mar 28 at 13:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @AkashShah I disagree: List<String> stringList = (List) numList; stringList.add("abc");. Lots of compiler warnings, but compiles and runs nevertheless.
  
  – Jiri Tousek
  Mar 28 at 22:44
  

  
  
  
  

 | 
show 3 more comments

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21

This is because of type erasure used with Java generics - the type checks are only performed at compile time for generic types, and the type info for generics is then erased, effectively turning List<Integer> into List<Object>.

My IDE warns you of an "Unchecked cast from Double to T". But the compiler couldn't be sure that your code is wrong, so it does not emit an error, just a warning.

Then at runtime, the type check is no longer present due to type erasure, so the code will run without error unless you perform some operation that fails due to incompatible runtime type. System.out.println() is not such operation.

---

If you change the print code to

Integer num = numList.get(0);
System.out.println(num);

this will now involve runtime type check and will therefore fail:

java.lang.ClassCastException: java.lang.Double cannot be cast to java.lang.Integer

share|improve this answer

edited Mar 28 at 12:05

answered Mar 28 at 11:55

Jiri TousekJiri Tousek

10.7k52240

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Note that a ClassCastException is emitted when one tries to do this: Integer i = numList.get(0).
  
  – MC Emperor
  Mar 28 at 11:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MCEmperor Thanks, added. I couldn't find a simple example to force type incompatibility - integer seems to basically have no meaningful methods that wouldn't either be static or already in Number.
  
  – Jiri Tousek
  Mar 28 at 12:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @JiriTousek You may be interested in Java is Unsound. Basically all java compilers will add cast checks in places to avoid issues because they know that the type system is broken.
  
  – Giacomo Alzetta
  Mar 28 at 12:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  effectively turning into List<Number> not List<Object> because of upper bound is Number. if its List<Object> then you can store String but you can't
  
  – Akash Shah
  Mar 28 at 13:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @AkashShah I disagree: List<String> stringList = (List) numList; stringList.add("abc");. Lots of compiler warnings, but compiles and runs nevertheless.
  
  – Jiri Tousek
  Mar 28 at 22:44
  

  
  
  
  

 | 
show 3 more comments

21

This is because of type erasure used with Java generics - the type checks are only performed at compile time for generic types, and the type info for generics is then erased, effectively turning List<Integer> into List<Object>.

My IDE warns you of an "Unchecked cast from Double to T". But the compiler couldn't be sure that your code is wrong, so it does not emit an error, just a warning.

Then at runtime, the type check is no longer present due to type erasure, so the code will run without error unless you perform some operation that fails due to incompatible runtime type. System.out.println() is not such operation.

---

If you change the print code to

Integer num = numList.get(0);
System.out.println(num);

this will now involve runtime type check and will therefore fail:

java.lang.ClassCastException: java.lang.Double cannot be cast to java.lang.Integer

share|improve this answer

edited Mar 28 at 12:05

answered Mar 28 at 11:55

Jiri TousekJiri Tousek

10.7k52240

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Note that a ClassCastException is emitted when one tries to do this: Integer i = numList.get(0).
  
  – MC Emperor
  Mar 28 at 11:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MCEmperor Thanks, added. I couldn't find a simple example to force type incompatibility - integer seems to basically have no meaningful methods that wouldn't either be static or already in Number.
  
  – Jiri Tousek
  Mar 28 at 12:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @JiriTousek You may be interested in Java is Unsound. Basically all java compilers will add cast checks in places to avoid issues because they know that the type system is broken.
  
  – Giacomo Alzetta
  Mar 28 at 12:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  effectively turning into List<Number> not List<Object> because of upper bound is Number. if its List<Object> then you can store String but you can't
  
  – Akash Shah
  Mar 28 at 13:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @AkashShah I disagree: List<String> stringList = (List) numList; stringList.add("abc");. Lots of compiler warnings, but compiles and runs nevertheless.
  
  – Jiri Tousek
  Mar 28 at 22:44
  

  
  
  
  

 | 
show 3 more comments

This is because of type erasure used with Java generics - the type checks are only performed at compile time for generic types, and the type info for generics is then erased, effectively turning List<Integer> into List<Object>.

My IDE warns you of an "Unchecked cast from Double to T". But the compiler couldn't be sure that your code is wrong, so it does not emit an error, just a warning.

Then at runtime, the type check is no longer present due to type erasure, so the code will run without error unless you perform some operation that fails due to incompatible runtime type. System.out.println() is not such operation.

---

If you change the print code to

Integer num = numList.get(0);
System.out.println(num);

this will now involve runtime type check and will therefore fail:

java.lang.ClassCastException: java.lang.Double cannot be cast to java.lang.Integer

share|improve this answer

edited Mar 28 at 12:05

answered Mar 28 at 11:55

Jiri TousekJiri Tousek

10.7k52240

Integer num = numList.get(0);
System.out.println(num);

share|improve this answer

edited Mar 28 at 12:05

answered Mar 28 at 11:55

Jiri TousekJiri Tousek

10.7k52240

answered Mar 28 at 11:55

Jiri TousekJiri Tousek

10.7k52240

answered Mar 28 at 11:55

Jiri TousekJiri Tousek

10.7k52240