如何在 Java 中使用 lambda 表达式实现 ToLongBiFunction<T, U>?
java 8object oriented programmingprogramming
ToLongBiFunction<T, U> 是 java.util.function 包中内置的函数式接口。该函数式接口接受 两个引用类型参数 作为输入，并返回一个 长值 结果。ToLongBiFunction<T, U> 接口可用作 lambda表达式 或 方法引用 的赋值目标，并且仅包含一个抽象方法:applyAsLong()。
语法
@FunctionalInterface
interface ToLongBiFunction<T, U> {
long applyAsLong(T t, U u);
}
示例
import java.util.*;
import java.util.function.ToLongBiFunction;
public class ToLongBiFunctionTest {
public static void main(String[] args) {
ToLongBiFunction<Map<String, Long>, String> getMobileNum = (map,value) -> { // lambda
if(map != null && !map.isEmpty()) {
if(map.containsKey(value)) {
return map.get(value);
}
}
return 0;
};
Map<String, Long> mobileNumbers = new HashMap<String, Long>();
mobileNumbers.put("Raja", 9959984805L);
mobileNumbers.put("Adithya", 7702144433L);
mobileNumbers.put("Jai", 7013536286L);
mobileNumbers.put("Chaitanya", 9652671506L);
String name = "Raja";
long result = getMobileNum.applyAsLong(mobileNumbers, name);
System.out.println("Raja's Mobile no: "+ result);
name = "Adithya";
System.out.println("Adithya's Mobile no: "+ getMobileNum.applyAsLong(mobileNumbers, name));
name = "Jai";
System.out.println("Jai's Mobile no: "+ getMobileNum.applyAsLong(mobileNumbers, name));
name = "Chaitanya";
System.out.println("Chaitanya's Mobile no: "+ getMobileNum.applyAsLong(mobileNumbers, name));
}
}
输出
Raja's Mobile no: 9959984805 Adithya's Mobile no: 7702144433 Jai's Mobile no: 7013536286 Chaitanya's Mobile no: 9652671506
