• [Functional Programming] Combine Multiple State ADT Instances with the Same Input (converge(liftA2(constant)))


    When combining multiple State ADT instances that depend on the same input, using chain can become quite burdensome. We end up having to play leapfrog with values inside of nested chain statements. Very reminiscent of the old callback nastiness we had to deal with before Promisesgraced our existence.

    But fear not, but taking advantage of the Applicative Functor portion of the State ADT in combination with the converge combinator, we can apply these types of transitions in unison, passing the value to both virtually simultaneously.

    For example:

    // () -> b
    const validateAnswer = converge(
        liftA2(equals),
        cardToHint,
        getHint
    )

    'validateAnswer' return a boolean value.

    We want to take this value, and pass to two functions 'setIsCorrect' & 'updateRank'

    // b -> ()
    const setIsCorrect = b => over('isCorrect', constant(b));
    
    // b -> ()
    const updateRank = b => over('rank', adjustRank(b));

    One functional way to do this is using 'converge':

    const applyFeedback = converge(
        liftA2(constant),
        setIsCorrect,
        updateRank
    )

    ----------------

    const {prop, State, omit, curry, converge,map, composeK, liftA2, equals, constant,option, chain, mapProps, find, propEq, isNumber, compose, safe} = require('crocks');
    const  {get, modify, of} = State; 
    
    const state = {
        cards: [
            {id: 'green-square', color: 'green', shape: 'square'},
            {id: 'orange-square', color: 'orange', shape: 'square'},
            {id: 'blue-triangle', color: 'blue', shape: 'triangle'}
        ],
        hint: {
            color: 'green',
            shape: 'square'
        },
        isCorrect: null,
        rank: 2
    }
    
    const inc = x => x + 1;
    const dec = x => x - 1;
    const incOrDec = b => b ? dec :  inc;
    const clamp = (min, max) => x => Math.min(Math.max(min, x), max);
    const clampAfter = curry((min, max, fn) => compose(clamp(min, max), fn));
    const limitRank = clampAfter(0, 4);
    const over = (key, fn) => modify(mapProps({[key]: fn}))
    const getState = key => get(prop(key));
    const liftState = fn => compose(
        of,
        fn
    )
    const getCard = id => getState('cards')
        .map(chain(find(propEq('id', id))))
        .map(option({}))
    const getHint = () => getState('hint')
        .map(option({}))
    const cardToHint = composeK(
        liftState(omit(['id'])),
        getCard
    )
    // () -> b
    const validateAnswer = converge(
        liftA2(equals),
        cardToHint,
        getHint
    )
    // b -> ()
    const setIsCorrect = b => over('isCorrect', constant(b));
    
    const adjustRank = compose(limitRank, incOrDec);
    // b -> ()
    const updateRank = b => over('rank', adjustRank(b));
    
    const applyFeedback = converge(
        liftA2(constant),
        setIsCorrect,
        updateRank
    )
    
    const feedback = composeK(
        applyFeedback,
        validateAnswer
    )
    
    console.log(
        feedback('green-square')
            .execWith(state)
    )
  • 相关阅读:
    少走弯路的10条忠告
    思考
    哈弗经典校训
    项目导出excel引发的一些问题
    hibernate 缓存设置
    dubbo简单用法
    sql 类型问题
    spring this.logger.isDebugEnabled()
    红黑树
    归并排序
  • 原文地址:https://www.cnblogs.com/Answer1215/p/10268647.html
Copyright © 2020-2023  润新知