CODEWITHSUNDEEP
javalambdajava-8
user1028880
user1028880

Reputation:

Java8 function type for `(a,b) -> a+b`

Trying to test Java8 lambda, but the type is confusing:

import java.util.function.ToIntBiFunction;
import java.util.stream.IntStream;

public class Test {
    public static void main(String... args) {

    int sum1 = 0;
    for (int n = 0; n < 10; n++) {
        sum1 += n;
       }


    ToIntBiFunction<Integer, Integer> add = (a, b) -> a + b;  
    int sum2 = IntStream.range(0, 10)
                        .reduce(0, add); //error here


     System.out.println(""+sum1);
     System.out.println(""+sum2);

   }
}

Test.java:15: error: incompatible types: ToIntBiFunction cannot be converted to IntBinaryOperator .reduce(0, add);

What is the most generic way to define the function

(a,b) -> a+b

Thanks.

Upvotes: 0

Views: 380

Answers (2)

Peter Lawrey
Peter Lawrey

Reputation: 533680

The most generic way is as a lambda, once you assign it to a variable, or cast it to a type it become a specific type.

Try the type the reduce() expects

IntBinaryOperator add = (a,b) -> a+b

or use the built in one.

int sum2 = IntStream.range(0, 10)
                    .reduce(0, Integer::sum);

Upvotes: 3

Konstantin Yovkov
Konstantin Yovkov

Reputation: 62864

Obviously, you need an IntBinaryOperator for the .reduce(), instead of ToIntBiFunction.

IntBinaryOperator add = (a, b) -> a + b;
int sum2 = IntStream.range(0, 10)
                    .reduce(0, add);

Upvotes: 0

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