java8中的lambda定义的函数该如何引用

Question

定义好的函数，不知道该如何使用。

// 不知道怎么引用
BinaryOperator<Long> add = (x, y) -> x + y;

Answer 1

public class Main {
    
    private long t, u;
    
    private Long test(BinaryOperator<Long> b) {
        return b.apply(t, u);
    }
    public static void main(String[] args) {
        
        Main m = new Main();
        m.t = 1; m.u = 2;
        BinaryOperator<Long> b = (x, y) -> x + y;
        System.out.println(m.test(b));
    }
}

It is useless to pull it out alone. Lambda only defines the operation method of data, that is, it defines a function. Specifically where to use it, you need to define a method with a lambda expression (functional interface) as the parameter, and then call the actual operation of lambda (which function in the interface definition) inside the method, such as accept.

Answer 2

Lambda定义的并不是函数，它只是匿名类的缩写方式，其生成的还是一个对象。就如你的例子中，它生成的一个BinaryOperator<Long>Object, then it is an instance object of this class. How to use it is the same as how to use an object.