What is the difference between <? extends Base> and <T extends Base>?

Question

In this example:

import java.util.*;  public class Example {     static void doesntCompile(Map<Integer, List<? extends Number>> map) {}     static <T extends Number> void compiles(Map<Integer, List<T>> map) {}      static void function(List<? extends Number> outer)     {         doesntCompile(new HashMap<Integer, List<Integer>>());         compiles(new HashMap<Integer, List<Integer>>());     } } 

doesntCompile() fails to compile with:

Example.java:9: error: incompatible types: HashMap<Integer,List<Integer>> cannot be converted to Map<Integer,List<? extends Number>>         doesntCompile(new HashMap<Integer, List<Integer>>());                       ^ 

while compiles() is accepted by the compiler.

This answer explains that the only difference is that unlike <? ...>, <T ...> lets you reference the type later, which doesn't seem to be the case.

What is the difference between <? extends Number> and <T extends Number> in this case and why doesn't the first compile?

Answer 1

By defining the method with the following signature:

static <T extends Number> void compiles(Map<Integer, List<T>> map) {} 

and invoking it like:

compiles(new HashMap<Integer, List<Integer>>()); 

you're matching T against the type you're providing.

In the jls §8.1.2 we find, that (interesting part bolded by me):

A generic class declaration defines a set of parameterized types (§4.5), one for each possible invocation of the type parameter section by type arguments. All of these parameterized types share the same class at run time.

In other words, the type T is matched against the input type and assigned Integer. The signature will effectively become static void compiles(Map<Integer, List<Integer>> map).

When it comes to doesntCompile method, jls defines rules of subtyping (§4.5.1, bolded by me):

A type argument T1 is said to contain another type argument T2, written T2 <= T1, if the set of types denoted by T2 is provably a subset of the set of types denoted by T1 under the reflexive and transitive closure of the following rules (where <: denotes subtyping (§4.10)):

• ? extends T <= ? extends S if T <: S

• ? extends T <= ?

• ? super T <= ? super S if S <: T

• ? super T <= ?

• ? super T <= ? extends Object

• T <= T

• T <= ? extends T

• T <= ? super T

This means, that ? extends Number indeed contains Integer or even List<? extends Number> contains List<Integer>, but it's not the case for Map<Integer, List<? extends Number>> and Map<Integer, List<Integer>>. More on that topic can be found in this SO thread. You can still make the version with ? wildcard work by declaring, that you expect a subtype of List<? extends Number>:

public class Example {     // now it compiles     static void doesntCompile(Map<Integer, ? extends List<? extends Number>> map) {}     static <T extends Number> void compiles(Map<Integer, List<T>> map) {}      public static void main(String[] args) {         doesntCompile(new HashMap<Integer, List<Integer>>());         compiles(new HashMap<Integer, List<Integer>>());     } } 

Answer 2

In the call:

compiles(new HashMap<Integer, List<Integer>>()); 

T is matched to Integer, so the type of the argument is a Map<Integer,List<Integer>>. It's not the case for the method doesntCompile: the type of the argument stays Map<Integer, List<? extends Number>> whatever the actual argument in the call; and that is not assignable from HashMap<Integer, List<Integer>>.

UPDATE

In the doesntCompile method, nothing prevents you to do something like this:

static void doesntCompile(Map<Integer, List<? extends Number>> map) {     map.put(1, new ArrayList<Double>()); } 

So obviously, it cannot accept a HashMap<Integer, List<Integer>> as the argument.