About Question enthuware.ocpjp.v8.2.1912 :

[ArpRokz]

Just an interesting thing I found and though I should share...

Even this works...

IntFunction<IntUnaryOperator> fo = a->b->a-b;

int x = (fo.apply(20)).applyAsInt(5);

System.out.println(15);

[saurabh.agarwal560]

I tried compiling it and found 15 as a result but I have a doubt that why it fo.apply(20) doesnt throw compilation error as fo is of type Intfunction and the function call should be fo.applyAsInt(20). Please explain.

[admin]

fo points to in IntFunction object, which has an apply() method not applyAsInt(). So, why do expect fo.applyAsInt() to work?
The internal implementation of this IntFunction returns an IntUnaryOperator. That is what the first post in this thread shows.