Typescript: declaring and implementing a curried Method

Question

In the skeleton of the following class

class Iterable<A> {

    filter( filter: ( value: A ) => boolean ): Iterable<A> { ... }

    map<B>( f: ( value: A, index?: number ) => B ): Iterable<B> { ...}

    collect = <B>( filter: ( value: A ) => boolean ) => ( mapper: ( value: A ) => B ): Iterable<B> => { ... }

}

there is a curried property collect which is implemented this way

collect = <B>( filter: ( value: A ) => boolean ) => ( mapper: ( value: A ) => B ): Iterable<B> => {
    return this.filter( filter ).map( mapper )
}

This property is basically a curried function that takes two functions as argument and returns an Iterable<B>. With this, being an Iterable<A>, it can be used in the following manner:

const it: Iterable<string> = this.collect<string>( f => true )( ( v: A ) => v.toString() )

Now, I would like to transform this property into a class method but I am failing to find the right syntax

declaring
```
collect<B>( filter: ( value: A ) => boolean ): ( mapper: ( value: A ) => B ) => Iterable<B>
```
will preserve its usage (see itabove), however I do not know how to implement it as the method expects to return a Function, not an Iterable

declaring

collect<B>( filter: ( value: A ) => boolean ) => ( mapper: ( value: A ) => B ) : Iterable<B>

is invalid...

Solution: (see @JLRishe and @NitzanTomer answers) I went for an explicit type declaration:

collect<B>(filter: (value: A) => boolean): (mapper: (value: A) => B) => Iterable<B> {
    return (mapper: (value: A) => B) => this.filter(filter).map(mapper);
}

Can you put the code for your class as well? For example what's this A, what does this.filter returns? etc.. — Nitzan Tomer
– Nitzan Tomer, Commented Mar 21, 2017 at 9:08
@NitzanTomer A is a generic, see my edits above with the skeleton of the class — Bruno Grieder
– Bruno Grieder, Commented Mar 21, 2017 at 9:20

JLRishe · Accepted Answer · 2017-03-21 09:26:17Z

1

You can implement it like this:

collect<B>(filter: (value: A) => boolean) {
    return (mapper: (value: A) => B) => this.filter(filter).map(mapper);
}

answered Mar 21, 2017 at 9:26

JLRishe

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Comments

Nitzan Tomer · Accepted Answer · 2017-03-21 09:27:16Z

1

How about

collect<B>(filter: (value: A) => boolean) {
    return (mapper: (value: A) => B): Iterable<B> => {
        return this.filter(filter).map(mapper);
    };
}

answered Mar 21, 2017 at 9:27

Nitzan Tomer

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1 Comment

Bruno Grieder Over a year ago

Thanks. Same solution as JLRishe... who has beaten you on the clock (according to SO)

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