Generic bounded wildcard of function input type

Question

While reading the Stream interface source code I found this method signature:

<R> Stream<R> map(Function<? super T, ? extends R> mapper);

I wonder why the input type of mapper is ? super T while the output type is ? extends R, why not use ? extends for both?

Answer 1

Consider you want to map a CharSequence to another CharSequence (thus T = R = CharSequence). Which functions will work for you?

Function<Object, String> fn1 = Object::toString;

Is it good for you? Yes, because it can take any CharSequence (which is also Object) and convert it to the String (which is also CharSequence).

Function<CharSequence, StringBuilder> fn2 = StringBuilder::new;

Is it good for you? Yes, because it can take any CharSequence and convert it to the StringBuilder (which is also CharSequence).

Function<String, String> fn3 = String::trim;

Is it good for you? No, because it cannot take any CharSequence, only some of them.

Thus you can see that first type argument must be CharSequence or any superclass, but second must be CharSequence or any subclass.

Answer 2

Because of PECS :) - Producer Extends, Consumer Super

Producer Extends - If you need a List to produce T values (you want to read Ts from the list), you need to declare it with ? extends T

Consumer Super - If you need a List to consume T values (you want to write Ts into the list), you need to declare it with ? super T