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

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 yesterday

gaby

asked yesterday

gabygaby

51912

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Possible duplicate of type erasure in implementation of ArrayList in Java
  
  – vaxquis
  yesterday
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  
  

  
  
  
  

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 yesterday

gaby

asked yesterday

gabygaby

51912

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Possible duplicate of type erasure in implementation of ArrayList in Java
  
  – vaxquis
  yesterday
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  
  

  
  
  
  

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 yesterday

gaby

asked yesterday

gabygaby

51912

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 yesterday

gaby

asked yesterday

gabygaby

51912

share|improve this question

edited yesterday

gaby

asked yesterday

gabygaby

51912

asked yesterday

gabygaby

51912

asked yesterday

gabygaby

51912

1 Answer
1

active

oldest

votes

20

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 yesterday

answered yesterday

Jiri TousekJiri Tousek

10.5k52240

• 
  
  
  

  
  3
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 3 more comments

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20

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 yesterday

answered yesterday

Jiri TousekJiri Tousek

10.5k52240

• 
  
  
  

  
  3
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 3 more comments

20

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 yesterday

answered yesterday

Jiri TousekJiri Tousek

10.5k52240

• 
  
  
  

  
  3
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

 | 
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 yesterday

answered yesterday

Jiri TousekJiri Tousek

10.5k52240

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

share|improve this answer

edited yesterday

answered yesterday

Jiri TousekJiri Tousek

10.5k52240

answered yesterday

Jiri TousekJiri Tousek

10.5k52240

answered yesterday

Jiri TousekJiri Tousek

10.5k52240