Map an objects type in typescript

Question

I am looking for a way of a "mapped" object type in typescript.

I have a the following typings:

interface Factory<T>{
     serialize: (val: T)=>void,
     deserialize: ()=>T,
}
interface MyDict{
     [key: string]: Factory<any>
}

function deserialize(dict: MyDict){
     let mapped = {};
     for(let k in dict){
          mapped[k] = dict[k].deserialize();
     }
     return mapped;
}

What I want is that the return type of map is correctly determined.

So when doing this:

let mapped = map({
    foo: {deserialize: ()=>'hello world'}, 
    foo2: {deserialize: ()=>123}, 
    foo3: {deserialize: ()=>({foo4: 'foo4'})}
});

mapped should be typed as {foo: string, foo2: number, foo3: {foo4: string}}.

Given MyDict is not generic (and map is not generic either) - it's unlikely you can have that. — zerkms
– zerkms, Commented Mar 13, 2020 at 20:29

Titian Cernicova-Dragomir · Accepted Answer · 2020-03-13 21:03:15Z

2

You can do this using a mapped type. The function will also need to be generic in order to capture the actual type of the argument:

interface Factory<T>{
     serialize?: (val: T)=>void,
     deserialize: ()=>T,
}
interface MyDict{
     [key: string]: Factory<any>
}

type FactoryReturnType<T extends MyDict> = {
    [K in keyof T]: ReturnType<T[K]['deserialize']>
}

function deserialize<T extends MyDict>(dict: T){
     let mapped = {} as FactoryReturnType<T>;;
     for(let k in dict){
          mapped[k] = dict[k].deserialize();
     }
     return mapped;
}

let mapped = deserialize({
    foo: {deserialize: ()=>'hello world'}, 
    foo2: {deserialize: ()=>123}, 
    foo3: {deserialize: ()=>({foo4: 'foo4'})}
});

mapped.foo3.foo4

Playground Link

edited Mar 13, 2020 at 21:03

answered Mar 13, 2020 at 20:39

Titian Cernicova-Dragomir

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3 Comments

lal12 Over a year ago

This almost seems to be the solution. However I had a bit of an unclear part in my question. Instead of just a "factory" function, I actually have a generic Interface with multiple properties derived from <T>, the factory or in my case deserialize function is just one of them. I updated my example accordingly.

Titian Cernicova-Dragomir Over a year ago

@LucaAdrian updated the answer. Sorry the initial code was not 100% consistent between call and definition so I took some decistions.

lal12 Over a year ago

Already tried that, just made a typo, but works wonderfully now ^^. So thanks a lot.

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