对浮点数求和的自定义收集器实现 java 8

Question

I am trying to create a custom float addition similar to Collectors.summingDouble() .

But I am facing 2 issues and I am not sure about how to fix it.

1. BiConsumer - Line #27 - void methods cannot return a value
2. Collectors.of - Line#32 - The method of(Supplier, BiConsumer<R,T> , BinaryOperator<R> , Collector.Characteristics...) in the type Collector is not applicable for the arguments ( Supplier<Float[]> , BiConsumer<Float,Employee> , BinaryOperator<Float> ) What needs to be done here for fixing the issue?

public class CustomCollector {


    public static void main(String[] args) {
        Employee e1=new Employee(1,"Tom",10.2f);
        Employee e2=new Employee(1,"Jack",10.4f);
        Employee e3=new Employee(1,"Harry",10.4f);
        ArrayList<Employee> lstEmployee=new ArrayList<Employee>();
        lstEmployee.add(e1);lstEmployee.add(e2);lstEmployee.add(e3);

/*  Implementation 1
 *  double totalSal=lstEmployee.stream().collect(Collectors.summingDouble(e->e.getSal()));
        System.out.println(totalSal);
*/  
        //Implementation 2
        Function<Employee,Float> fun=(e)->e.getSal();
        BiConsumer<Float,Employee> consumer=(val,e)->val+e.getSal();
        BinaryOperator<Float> operator=(val1,val2)->val1+val2;
        Supplier<Float[]> supplier=() -> new Float[2];

        float FtotalSal=lstEmployee.stream().collect(
                Collector.of(supplier,consumer,operator));
        System.out.println(FtotalSal);
    }

}

class Employee {
    int id;
    String name;
    float sal;
    // getters, setter, constructror
}

Answer 1

It seems, you are confusing Reduction with Mutable Reduction .

Your functions, (val, e) -> val + e.getSal() and (val1, val2) -> val1 + val2) , are suitable for a Reduction operation, but not for collect . The supplier, producing a Float[] array of length two, doesn't fit to it at all.

Eg, you can perform your operation using

float f = lstEmployee.stream().reduce(
    0F,
    (val, e) -> val + e.getSal(),
    (val1, val2) -> val1 + val2);

This bears some boxing overhead, as all intermediate sums are represented as Float objects.

You can avoid this using Mutable Reduction , when you create a mutable container which is capable of holding a float value without boxing, ie new float[1] . Then, you have to supply functions accepting an array argument and changing the contained value. Since your intended result is a float , rather than an array, you further need a finisher to produce the final result.

float f = lstEmployee.stream().collect(
    Collector.of(
        () -> new float[1], // a container capable of holding one float
        (floatArray,e) -> floatArray[0] += e.getSal(), // add one element to the array
        (a1, a2) -> { a1[0] += a2[0]; return a1; }, // merge two arrays
        array -> array[0]) // extracting the final result value
    );

Of course, this is only for exercising, as you have already shown to know a simpler solution using builtin features.

Answer 2

If getting the sum is the focus. You could try something like below.

 float FtotalSal = (float) lstEmployee.stream().mapToDouble(e -> e.getSal()).sum();