Java:将 lambda 存储为值的映射

Question

我想学习 Java 的“较新”语法和 API 中的可能性。我所说的较新是指 10+(比如说 10-13)。它主要围绕 lambda 声明并存储符合与映射中的值相同签名的不同实现。由于最近我主要与 Gosu 合作，我可以提供以下片段:

var longInput = 10000000L

new LinkedHashMap<String, block(long) : long>( {
    "byte"  -> (\x -> x as byte as long),
    "short" -> (\x -> x as short as long),
    "int"   -> (\x -> x as int as long),
    "long"  -> (\x -> x as long as long)
}).eachKeyAndValue(\key, value ->
  print("${longInput} ${value(longInput) == longInput ? "can" : "cannot"} be converted to ${key}")
)

我可以在 Java 10 中进行类似的操作:

import java.util.*;

public class Test {
    public static void main(String[] args) {
        long longInput = 10000000L;

        var conversions = new LinkedHashMap<String, Conversion<Long>>();
        conversions.put("byte",  (x) -> (long) (byte)  x.longValue());
        conversions.put("short", (x) -> (long) (short) x.longValue());
        conversions.put("int",   (x) -> (long) (int)   x.longValue());
        conversions.put("long",  (x) -> (long) (long)  x.longValue());

        conversions.forEach((key, value) -> {
            System.out.printf("%d %s be converted to %s%n", longInput, value.convert(longInput) == longInput ? "can" : "cannot", key);      
        });
    }
}

interface Conversion<T> {
    T convert(T input);
}

我的问题:

1. 是否可以在没有命名接口(interface)的情况下通过类似 Gosu 中的“匿名”函数方式来完成？
2. 还有什么可以让 Java 中的代码更加简洁吗？

更新:这只是一些围绕代码的游戏，其目的是将原始 long 类型双重转换为较小类型并返回。灵感来自https://www.hackerrank.com/challenges/java-datatypes/problem 。所以从我的角度来看，我想留下来。

使用答案，我当前的 Java 10 代码将如下所示:

public class Test { 
    public static void main(String[] args) {
        var longInput = 10000000L;

        new LinkedHashMap<String, UnaryOperator<Long>>() {{
            put("byte",  (x) -> (long) (byte)  x.longValue());
            put("short", (x) -> (long) (short) x.longValue());
            put("int",   (x) -> (long) (int)   x.longValue());
            put("long",  (x) -> (long) (long)  x.longValue());
        }}.forEach((key, value) -> {
            System.out.printf("%d %s be converted to %s%n", longInput, value.apply(longInput) == longInput ? "can" : "cannot", key);        
        });
    }
}

Answer 1

Could it be done without having a named interface but in similar 'anonymous' function way like in Gosu?

Java 已经有一个与您定义的类似的FunctionalInterface。您可以使用UnaryOperator<Long>您的值(value)观Map .

Anything else that could make this more concise in Java?

我认为这读起来更好:

Map<String, UnaryOperator<Long>> conversions = new LinkedHashMap<>();
conversions.put("byte", a -> (long) a.byteValue());
conversions.put("short", a -> (long) a.shortValue());
conversions.put("int", a -> (long) a.intValue());
conversions.put("long", a -> a); // UnaryOperator.identity()