TypeScript generic use default value only

Question

I'm trying to apply a default value to a generic. But whenever the generic is not provided, the generic is defaulting to the value given to an argument. I want to prevent this from happening. Here's a very simple example:

type Data = Record<string, any>

const computeData = <D extends Data = {}>(input: D): D => {
  // Does some stuff to input
  return input;
}

const res = computeData<{ name: string }>({ name: 'john' })

In this case the type of "res" is { name: string } as expected. But if the generic is not provided, I need it to default to {} which is the type given as the default value for the generic.

const computeData = <D extends Data = {}>(input: D): D => {
  // Does some stuff to input
  return input;
}

const res = computeData({ name: 'john' })

I have simplified the code for this example. I know in the context of this example this is kind of pointless.

EDIT: The actual use case is a network mocker that lets users add objects to mock graphql requests. I want to give them the option to provide resolver types. However, if they do not provide the types they are inferred from initial data passed in - I want to avoid this.

The class defining a generic

Type error because the initial call to constructor infers type for generic

This works because an explicit type is given when instantiating the class

this shows {name: string} for both of them: typescriptlang.org/play?#code/… — tevemadar
– tevemadar, Commented Jun 3, 2021 at 16:08
This is going to be very difficult thing to make Typescript do. You're basically breaking how generic functions infer their types. It may be possible, but it's not likely to be pretty. I'm really curious what the actual use case is here, as I suspect there's a better way. — Alex Wayne
– Alex Wayne, Commented Jun 3, 2021 at 16:43
I agree @AlexWayne, I suspect using a required type argument is more TypeScript-y than trying to work around the language, and my solution (posted as an answer) isn't pretty, but it is totally possible. — evelynhathaway
– evelynhathaway, Commented Jun 3, 2021 at 16:57
@AlexWayne thanks for the message. I have edited the post to include the actual use case which hopefully explains more realistically what I am hoping to achieve. The conditional types seem like a workable solution though. — wazzaday
– wazzaday, Commented Jun 4, 2021 at 11:23

evelynhathaway · Accepted Answer · 2021-06-03 17:13:11Z

4

You can get around the type inference by using another type parameter and filtering out the default values. You can even do this without having to make type assertions.

Example Code

type Data = Record<string, any>

type NoInferData<DataType extends Data | void, InputType extends Data> = (
    DataType extends void
        // If it was void, there was no argument, so the type must be an empty object, or whatever you'd like
        ? Record<string, never>
        // Otherwise it should match the explcitly added data type
        : InputType
); 

const computeData = <
    // Type arugment for the type of Data you are providing
    DataType extends Data | void = void,
    // Type argument to store the inferred type of the input
    // - If its void, return the default type, in this example it's an empty object
    // - Otherwise, return the type of Data explicitly provided
    InputType extends Data = DataType extends void ? Record<string, never> : DataType
>(input: NoInferData<DataType, InputType>): NoInferData<DataType, InputType> => {
  // Does some stuff to input
  return input;
}


// Allows expliclty marking the data type
const explicitData = computeData<{ name: string }>({ name: 'john' })

// Doesn't allow data that doesn't match the explcit data type
const explicitDataError = computeData<{ pizza: string }>({ name: 'john' })

// Doesn't allow data that doesn't meet the default (no inferring)
const implictEmptyObjectError = computeData({ name: 'john' })

// Allows the default type of the empty object
const implictEmptyObject = computeData({})

TypeScript Playground

edited Jun 3, 2021 at 17:13

answered Jun 3, 2021 at 16:55

evelynhathaway

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2 Comments

wazzaday Over a year ago

thanks for the reply. This looks like it could work. I'll see if I can get it working.

wazzaday Over a year ago

Got this working but as you say it's not pretty. Maybe the best course as you mention is to always require something const mockNetwork2 = new MockNetwork<Record<string, any>>()

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