How to use generic function interface method type parameters in a lambda

Question

Im trying to create a generic lambda so I dont need to redefine it for individual types. To do that I need to access the type parameters that were passed to the function. Unfortunately I didnt find any documentation on how to do that.

This is an example of what I want to do but the in front of the lambda wont compile:

import java.util.ArrayList;

public class Test {

    interface SolveFunction {
        <In, Out> Out apply(In in);
    }

    public static void main(String...args) {
        SolveFunction f = <In, Out> (In a)  -> {
            ArrayList<Out> list = new ArrayList<>();
            return list;
        };
        System.out.println(f.<String, Integer>apply("Hi").size());
    }
}

What are you trying to do exactly? There are many things in this code that won't compile... — zetches
– zetches, Commented Nov 19, 2013 at 13:55
Im trying to implement the Divide and Conquer Pattern which accepts 4 functions: Divide, Solve, Trivial and Combine, heres the larger piece of code why I want generics inside the lambda dpaste.de/R6ew#L76 (code not tested yet since it breaks when at the generics when i try to compile it) — user1703761
– user1703761, Commented Nov 19, 2013 at 13:58
FYI this is not production code and I know that Java is not inteded to be used as a functional language. — user1703761
– user1703761, Commented Nov 19, 2013 at 14:05

Jesper · Accepted Answer · 2013-11-19 14:23:54Z

7

First of all, add the type arguments to the interface itself:

interface SolveFunction<In, Out> {
    Out apply(In in);
}

You could implement a method:

public static <In, Out> List<Out> makeList(In in) {
    return new ArrayList<Out>();
}

Which you can then use as the implementation of the lambda:

SolveFunction<String, Integer> f = Test::makeList;

System.out.println(f.apply("Hi").size());

answered Nov 19, 2013 at 14:23

Jesper

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1 Comment

Jesper Over a year ago

Note also that you wouldn't even need your SolveFunction interface, there's already an interface in the standard library that looks the same: java.util.function.Function<T, R>.

zetches · Accepted Answer · 2013-11-19 14:27:13Z

0

I'm unsure about what you're trying to use your ArrayList for but you need to add the type declaration to the interface as well. The following example will compile:

public class Test {

    interface SolveFunction<In, Out> {
        Out apply(In in);
    }

    public static void main(String... args) {
        SolveFunction<String, Integer> f = new SolveFunction<String, Integer>() {

            @Override
            public Integer apply(String in) {
                return in.length();
            }
        };

        System.out.println(f.apply("Hi")); // Prints 2
        System.out.println(f.apply("HELLO")); // Prints 5
    }
}

answered Nov 19, 2013 at 14:27

zetches

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1 Comment

Jesper Over a year ago

But it doesn't use Java 8 lambda syntax and the method uses concrete types (String and Integer) and not type parameters.

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