Function Interface
Guys i am trying to figure out functional interfaces , i am learning how to use public interface Function<T, R> apply method it is easy. we can acsess to apply method wit lambda expression or anonymouse class. like
here interface instances doubleFunction and squareFunction .apply method will return T which in ourcase is Integer value;
now where it becomes bit confusing i wanted to test compose method
creating new instance of Function interface but then it is very confusing i called compose method with my squareFunction it should take doubleFunction function as a parameter and return <V> Function<V, R>
this V R T are very confusing
also why does that work ? how we apply value with composeFunction instance?
Function<Integer, Integer> doubleFunction = new Function<Integer, Integer>() {
@Override
public Integer apply(Integer integer) {
return integer * 2;
}
};
Function<Integer, Integer> squareFunction = new Function<Integer, Integer>() {
@Override
public Integer apply(Integer integer) {
return integer * integer;
}
};
here interface instances doubleFunction and squareFunction .apply method will return T which in ourcase is Integer value;
now where it becomes bit confusing i wanted to test compose method
Function<Integer, Integer> composeFunction = squareFunction.compose(doubleFunction);
creating new instance of Function interface but then it is very confusing i called compose method with my squareFunction it should take doubleFunction function as a parameter and return <V> Function<V, R>
this V R T are very confusing
int result = composeFunction.apply(5);
also why does that work ? how we apply value with composeFunction instance?
