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

Java generics type erasure: when and what happens?type erasure in implementation of ArrayList in JavaHow do I efficiently iterate over each entry in a Java Map?How do I call one constructor from another 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?How do I determine whether an array contains a particular value in Java?How do I declare and initialize an array in Java?How to split a string in JavaConverting 'ArrayList<String> to 'String[]' in JavaHow do I convert a String to an int in Java?

12

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 19 mins ago

gaby

asked 1 hour ago

gabygaby

48411

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of type erasure in implementation of ArrayList in Java
  
  – vaxquis
  18 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  17 mins ago
  

  
  
  
  

add a comment | 

12

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 19 mins ago

gaby

asked 1 hour ago

gabygaby

48411

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of type erasure in implementation of ArrayList in Java
  
  – vaxquis
  18 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  17 mins ago
  

  
  
  
  

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 19 mins ago

gaby

asked 1 hour ago

gabygaby

48411

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 19 mins ago

gaby

asked 1 hour ago

gabygaby

48411

share|improve this question

edited 19 mins ago

gaby

asked 1 hour ago

gabygaby

48411

asked 1 hour ago

gabygaby

48411

asked 1 hour ago

gabygaby

48411

1 Answer
1

active

oldest

votes

14

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 1 hour ago

answered 1 hour ago

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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  45 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  37 mins ago
  
  
  

  
  
  
  

add a comment | 

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14

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 1 hour ago

answered 1 hour ago

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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  45 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  37 mins ago
  
  
  

  
  
  
  

add a comment | 

14

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 1 hour ago

answered 1 hour ago

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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  45 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  37 mins ago
  
  
  

  
  
  
  

add a comment | 

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 1 hour ago

answered 1 hour ago

Jiri TousekJiri Tousek

10.5k52240

Integer num = numList.get(0);

System.out.println(num);

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Jiri TousekJiri Tousek

10.5k52240

answered 1 hour ago

Jiri TousekJiri Tousek

10.5k52240

answered 1 hour ago

Jiri TousekJiri Tousek

10.5k52240