typescript fp-ts如何使用函数来咖喱选项.fold

wfypjpf4 于 2023-02-05 发布在 TypeScript

关注(0)|答案(1)|浏览(130)

我想用这个形状创建一个函数来折叠一个选项

(option<A>) => ( fun: (<A>) => <B> ) => <B>

目前的代码看起来是这样的，但是我感觉有更惯用的方法来实现相同的效果

const stepIntrD = fu.pipe(
    lectureInteractionStats.children,
    option.fromNullable,
    option.chain(readonlyArray.findFirst<InteractionStatsData>((lis) => lis.id == step.id)),
    (v) => (fun: (_: InteractionStatsData) => boolean) => option.fold(() => false, fun)(v)
);
        
const wazVisited = stepIntrD(wasVisited);

typescript

来源：https://stackoverflow.com/questions/75295747/fp-ts-how-to-curry-option-fold-with-function

1条答案

按热度按时间

brccelvz1#

下面是编写代码的其他一些更惯用的方法：

const stepIntrD = (fun: (_: InteractionStatsData) => boolean): boolean => fu.pipe(
    lectureInteractionStats.children,
    option.fromNullable,
    option.chain(readonlyArray.findFirst<InteractionStatsData>((lis) => lis.id == step.id)),
    option.fold(() => false, fun)
);

// This version evaluates interactionStatsData once, which is closer to your original code,
// whereas the version above evaluates that code every time stepIntrD is run
const interactionStatsData = fu.pipe(
    lectureInteractionStats.children,
    option.fromNullable,
    option.chain(readonlyArray.findFirst<InteractionStatsData>((lis) => lis.id == step.id))
);
const stepIntrD = (fun: (_: InteractionStatsData) => boolean): boolean => fu.pipe(
    interactionStatsData,
    option.fold(() => false, fun)
);

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typescript fp-ts如何使用函数来咖喱选项.fold

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