Is it possible to return another function in a lambda expression in java? Say I have the following lambda expressions
addFive = (x) -> x+5;
addFiveIfPositive = (y) -> {
if (y > 0) return addFive;
else return y;
};
To give some context behind I'm asking this, I'm working with lambda expressions to define Comparators and for some comparators, I want to 'chain' comparators that I've already defined to not write the same code again, given some conditions like in the example above.
Any help would be appreciated.
Edit: This is a more representative example of what I'm working with.
import java.util.Comparator;
class Random{
Integer val;
String str;
Random(int val, String str){
this.val = new Integer(val);
this.str = str;
}
}
public class Temp{
static Comparator<Random> c1 = (r1, r2) -> (r1.val).compareTo((r2.val));
static Comparator<Random> c2 = (r1, r2) -> {
if ((r1.str).compareTo((r2.str)) == 0){
return c1.apply(r1,r2);
}
return (r1.str).compareTo((r2.str));
};
public static void main(String args){
Random rand1 = new Random(1, "Hello");
Random rand2 = new Random(2, "Hello");
System.out.println(c2.compare(rand1, rand2));
}
}
Using .apply() gives me the error
Temp.java:18: error: cannot find symbol
return c1.apply(r1,r2);
^
symbol: method apply(Random,Random)
location: variable c1 of type Comparator<Random>
According to the edit, all you are looking for is to make use of:
Comparator<Random> c1 = Comparator.comparing(Random::getVal);
Comparator<Random> c2 = Comparator.comparing(Random::getStr).thenComparing(c1);;
that is because you already have Comparators supporting the use case of chaining with the thenComparing API. Do note that this is just a simplification of the following code:
Comparator<Random> c1 = (r1, r2) -> (r1.val).compareTo((r2.val));
Comparator<Random> c2 = (r1, r2) -> {
if ((r1.str).compareTo((r2.str)) == 0){
return c1.compare(r1,r2); //'compare' instead of 'apply'
}
return (r1.str).compareTo((r2.str));
};
Prior to the edit, you could have used a ternary operator to represent the condition such as:
IntFunction<Integer> addFive = x -> x + 5;
IntFunction<Integer> addFiveIfPositive = y -> y > 0 ? y + 5 : y;