FP技巧

Posted 2020-11-24 cherrytab

柯里化

在理论计算机科学中，柯里化提供了在简单的理论模型中，比如：只接受一个单一参数的lambda演算中，研究带有多个参数的函数的方式。

 

实际上就是单可变参数的情况下我们进行一次逻辑抽取，这是模块化代码的一个重要思路。

package lambdasinaction.chap14;

import java.util.function.DoubleUnaryOperator;


public class Currying {

    public static void main(String[] args) {
        DoubleUnaryOperator convertCtoF = curriedConverter(9.0/5, 32);
        DoubleUnaryOperator convertUSDtoGBP = curriedConverter(0.6, 0);
        DoubleUnaryOperator convertKmtoMi = curriedConverter(0.6214, 0);

        System.out.println(convertCtoF.applyAsDouble(24));
        System.out.println(convertUSDtoGBP.applyAsDouble(100));
        System.out.println(convertKmtoMi.applyAsDouble(20));

        DoubleUnaryOperator convertFtoC = expandedCurriedConverter(-32, 5.0/9, 0);
        System.out.println(convertFtoC.applyAsDouble(98.6));
    }

    static double converter(double x, double y, double z) {
        return x * y + z;
    }

    static DoubleUnaryOperator curriedConverter(double y, double z) {
        return (double x) -> x * y + z;
    }

    static DoubleUnaryOperator expandedCurriedConverter(double w, double y, double z) {
        return (double x) -> (x + w) * y + z;
    }
}

 

持续数据化，数据透明原则

package lambdasinaction.chap14;

import java.util.function.Consumer;

public class PersistentTrainJourney {

    public static void main(String[] args) {
        TrainJourney tj1 = new TrainJourney(40, new TrainJourney(30, null));
        TrainJourney tj2 = new TrainJourney(20, new TrainJourney(50, null));

        TrainJourney appended = append(tj1, tj2);
        visit(appended, tj -> { System.out.print(tj.price + " - "); });
        System.out.println();

        // A new TrainJourney is created without altering tj1 and tj2.
        TrainJourney appended2 = append(tj1, tj2);
        visit(appended2, tj -> { System.out.print(tj.price + " - "); });
        System.out.println();

        // tj1 is altered but it‘s still not visible in the results.
        TrainJourney linked = link(tj1, tj2);
        visit(linked, tj -> { System.out.print(tj.price + " - "); });
        System.out.println();

        // ... but here, if this code is uncommented, tj2 will be appended
        // at the end of the already altered tj1. This will cause a
        // StackOverflowError from the endless visit() recursive calls on
        // the tj2 part of the twice altered tj1.
        /*TrainJourney linked2 = link(tj1, tj2);
        visit(linked2, tj -> { System.out.print(tj.price + " - "); });
        System.out.println();*/
    }

    static class TrainJourney {
        public int price;
        public TrainJourney onward;

        public TrainJourney(int p, TrainJourney t) {
            price = p;
            onward = t;
        }
    }

    static TrainJourney link(TrainJourney a, TrainJourney b) {
        if (a == null) {
            return b;
        }
        TrainJourney t = a;
        while (t.onward != null) {
            t = t.onward;
        }
        t.onward = b;
        return a;
    }

    static TrainJourney append(TrainJourney a, TrainJourney b) {
        return a == null ? b : new TrainJourney(a.price, append(a.onward, b));
    }

    static void visit(TrainJourney journey, Consumer<TrainJourney> c) {
        if (journey != null) {
            c.accept(journey);
            visit(journey.onward, c);
        }
    }
}

 

两个比较使用的例子