Collections.sort() declaration: why <? super T> rather than <T>

Question

Why does Collections.sort(List<T>) have the signature :

public static <T extends Comparable<? super T>> void sort(List<T> list)  

and not :

public static <T extends Comparable<T>> void sort(List<? extends T> list) 

• I understand that they both would serve the same purpose; so why did the framework developers use the first option?
• Or are these declarations really different?

Answer 1

Your proposed signature would probably work in Java-8. However in previous Java versions type inference was not so smart. Consider that you have List<java.sql.Date>. Note that java.sql.Date extends java.util.Date which implements Comparable<java.util.Date>. When you compile

List<java.sql.Date> list = new ArrayList<>(); Collections.sort(list); 

It perfectly works in Java-7. Here T is inferred to be java.sql.Date which is actually Comparable<java.util.Date> which is Comparable<? super java.sql.Date>. However let's try your signature:

public static <T extends Comparable<T>> void sort(List<? extends T> list) {}  List<java.sql.Date> list = new ArrayList<>(); sort(list); 

Here T should be inferred as java.util.Date. However Java 7 specification does not allow such inference. Hence this code can be compiled with Java-8, but fails when compiled under Java-7:

Main.java:14: error: method sort in class Main cannot be applied to given types;         sort(list);         ^   required: List<? extends T>   found: List<Date>   reason: inferred type does not conform to declared bound(s)     inferred: Date     bound(s): Comparable<Date>   where T is a type-variable:     T extends Comparable<T> declared in method <T>sort(List<? extends T>) 1 error 

Type inference was greatly improved in Java-8. Separate JLS chapter 18 is dedicated to it now, while in Java-7 the rules were much simpler.