Using a parameter to define the type of another parameter in the same function in TypeScript

user_19010015 0
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I have a function that receives two parameters, let's call them X and Y, and a type Z that defines a simple object. I want to know if there is a way to use the first parameter, X, to access the type of Z to dynamically define the correct type that the second argument must be in reason of X.

Some code bellow to illustrate:

type MyType = {
    key1: string,
    key2: number,
    key3: Array<number>
}

// I know that I can't assigned the y type like bellow, but's it's just to illutraste what I want
function myFunction(x: keyof MyType, y: MyType[x]) {
    // function logic
    // ...
}

// Then, i wan't the following behavior

myFunction('key1', 'some string');  // It compiles/works
myFunction('key1', 1);  // It gives error
myFunction('key2', 1);  // It compiles/works
myFunction('key3', 'another string');  // It gives error

Is the effect that I want possible to achieve? If so, how?

typescript
1 Answers

Yes, you can do that by giving the function two generic type parameters (I've called them KeyType and ValueType) where you use KeyType in the definition of ValueType, like this:

function myFunction<
    KeyType extends keyof MyType,
    ValueType extends MyType[KeyType]
>(x: KeyType, y: ValueType) {
    // function logic
    // ...
}

You don't have to provide explicit type arguments for them when you call the function, TypeScript can usually infer them from usage, so all of your usage examples work exactly as you want without being updated.

Playground link

user_1572470
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