Java 8: Syntax for Implementing java.util.function.Function as a Lambda Expression

Question

I'm having trouble with Java 8's syntax for lambda expressions, I can't seem to create one on the fly in one line. The following code works,

Function<Integer, Integer> increment = (x) -> (x + 1);
doStuff(increment);

but the following lines do not

doStuff((x) -> (x + 1));
doStuff(Function (x) -> (x + 1));
doStuff(new Function (x) -> (x + 1));
doStuff(Function<Integer, Integer> (x) -> (x + 1));
doStuff(new Function(Integer, Integer> (x) -> (x + 1));
doStuff(new Function<Integer, Integer>(x -> {x + 1;}));

and I'm not quite sure what else I can try. I certainly don't want to use

doStuff(new Function<Integer, Integer>() {
    @Override
    public Integer apply(Integer x){
        return x + 1;
    }
});

so what else is there? I've looked through a bunch of questions on lambda expression syntax and nothing seems to be working.

Community · Accepted Answer · 2017-05-23 12:13:10Z

4

Simply

doStuff((x) -> (x + 1));

You had

Function<Integer, Integer> increment = (x) -> (x + 1);
doStuff(increment);

So just replace it with the right hand side of = (generally).

(x) -> (x + 1)

If doStuff's single parameter is not of type Function<Integer, Integer>, you'll need a target functional interface type

doStuff((Function<Integer,Integer>) (x) -> (x + 1));

Your method was using a raw Function type. Read

What is a raw type and why shouldn't we use it?

edited May 23, 2017 at 12:13

CommunityBot

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answered Dec 7, 2014 at 0:25

Sotirios Delimanolis

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5 Comments

user3002473 Over a year ago

Does not work either, doStuff requires a Function.

Sotirios Delimanolis Over a year ago

@user3002473 Please explain what errors you are getting. Please give the signature of doStuff.

user3002473 Over a year ago

When I leave is as (x) -> (x + 1), it gives me the error bad operand types for binary operator '+'. If I put (Integer x) -> (x + 1), it says no suitable method found for doStuff((int x)->(x + 1)).

user3002473 Over a year ago

The signature of doStuff is doStuff(Function function).

Sotirios Delimanolis Over a year ago

@user3002473 It's raw. Parameterize it as Function<Integer, Integer> or do what I added in my most recent edit.

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