Chaining Functional Interfaces - IntUnaryOperator vs UnaryOperator

Question

I'm still learning functional interfaces. I'd like to know why I'm able to chain a

UnaryOperator

to the end of a

Function

, but not an

IntUnaryOperator

to the end of the same Function.

UnaryOperator <String> twoOfMe = s -> s + s;

Function <String, Integer> convertMe = s -> Integer.parseInt (s);



UnaryOperator <Integer> twiceMe = n -> 2*n;



IntUnaryOperator doubleMe = n -> 2*n;



int a = twoOfMe.andThen(convertMe).andThen(twiceMe).apply ("2");



int b = twoOfMe.andThen(convertMe).andThen(doubleMe).apply ("2");

int a

works with

twiceMe

but

int b

doesn't work with the

doubleMe

.

Thanks

Edit:
It says incompatible types. Required int. Found java.lang.Object

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Answer 1

andThen(Function<? super R, ? extends V> after) expects a Function argument. UnaryOperator<Integer> is a sub-interface of Function<Integer,Integer>, which matches. IntUnaryOperator has no relation to the Function interface, so doubleMe cannot be passed to andThen.

Chaining Functional Interfaces - IntUnaryOperator vs UnaryOperator

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