I have a very basic question ragarding Java generics . I thought that both List<Number> and List<? extends Number> are homogeneous . Am I right or is there something fundamental I am missing ?
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Generic types are more pedantic.
<? extends Number> means Number or an unknown a sub class. If you obtain such a value it will be a Number, but you cannot give a value of this type because you don't know which is valid.
The difference is in arguments and return values.
List<Number> numbers = new ArrayList<Number>();
Number n = 1;
numbers.add(n); // ok.
n = numbers.get(0); // ok
numbers.add(1); // ok.
List<? extends Number> numbers2 = new ArrayList<Double>();
numbers2.add(n); // not ok
n = numbers2.get(0); // ok
List<? super Number> numbers3 = new ArrayList<Serializable>();
numbers3.add(n); // ok
n = numbers3.get(0); // not ok.
super is used in a few places to signify the type can be a super type. e.g.
In Collections, this method says the Comparator needs to be able to compare the same type or any super type.
public static <T> void sort(List<T> list, Comparator<? super T> c)
This means you can have
Comparator<Number> comparesAnyNumbers = ...
List<Integer> ints = ...
Collections.sort(ints, comparesAnyNumbers);
Peter Lawrey
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They are more specific, `
` means just Number not some other type as well. – Peter Lawrey Aug 31 '12 at 13:34 -
Could you provides details on the usefullness of `List super Number>` ? – gontard Aug 31 '12 at 13:48
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1IMHO Its not terribly useful. ;) – Peter Lawrey Aug 31 '12 at 13:52
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@PeterLawrey getting compilation issue for `numbers2.add(new Integer(2)); `? this should work right? – Nandkumar Tekale Aug 31 '12 at 13:52
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@Nandkumar No, numbers2 is actually an `ArrayList
` – Peter Lawrey Aug 31 '12 at 13:56 -
@PeterLawrey : I used diamond `<>` there, no
. – Nandkumar Tekale Aug 31 '12 at 13:58 -
1@PeterLawrey from the compiler perspective, number2 is a `List extends Number>` not a `ArrayList
`. The instruction `numbers2.add(Double.valueOf(1d));` don't work too. – gontard Aug 31 '12 at 13:59 -
@gontard Correct, the point is; the list contains a unknown type which extends Number. – Peter Lawrey Aug 31 '12 at 14:02
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1@PeterLawrey nothing (but `null`) could be added into `List extends Number>` ? – gontard Aug 31 '12 at 14:04
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`List extends T>` : since we don't what is really `?` : you can't **add** item into it : this is a sort of `"read-only compile time checked list"`. – gontard Aug 31 '12 at 14:17
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`List super T>` : since we don't what is really `?` : you can't **read** item from it : this is a sort of `"write-only compile time checked list"`. – gontard Aug 31 '12 at 14:19
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@gontard Sort of but you can call `remove(0);` or `clear();` on it. ;) – Peter Lawrey Aug 31 '12 at 14:19
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@gontard You can read an `Object` from `? super T` as every object is an `Object` – Peter Lawrey Aug 31 '12 at 14:20
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@gontard `const` is a keyword in Java, you just can't use it anywhere. :D – Peter Lawrey Aug 31 '12 at 14:23
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@PeterLawrey may be in `java 10` ;) – gontard Aug 31 '12 at 14:25
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Generics are compile time language features, means, they don't exist in run-time. In generic mechanism, for compile-time checks, they are not homogeneous, i.e. if you want to use polymorphism in generic type.
Following gives you a compile time error, although it seems a valid definition:
List<Number> list = new ArrayList <Integer>();
whereas
List<? extends Number> list = new ArrayList <Integer>();
is valid. Moreover, you can't use wildcard types on the right side:
List list = new ArrayList <? extends Integer>();
won't be compiled.
Erhan Bagdemir
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List<Number> --> List of Numbers (or instances of Number)
List<? extends Number> --> List any type which extends Number
Santosh
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