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 ? super T while the output type is ? extends R ? extends R , why not use ? extends ? 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 ? 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 ? super T

Answer 3

There is a rule of thumb which was first mentionned by Joshua Bloch in his Effective Java book which stands as PECS .

• PE Producer extends
• CS Consumer super

This function produce a result (? extends R) from what it received ( ? super T ).

The documentation of java.util.Function is kinda clear on that :

Represents a function that accepts one argument and produces a result.