Typescript: force key and value to be of the same type?

Question

I'm trying to create a type to express the following constraint:

// Valid
{
  "foo": "foo"
}

// Valid
{
  "bar": "bar"
}

// Not valid
{
  "foo": "bar"
}

I think I should be able to do this with mapped types:

type KeyValueSameDict = {
  [key in string]: key
};

However, this simply infers the type Record<string, string>. I also think I could do this if I knew the keys ahead of time:

type KeyValueSameDictKeys = "foo" | "bar";

type KeyValueSameDict = {
  [v in KeyValueSameDictKeys]?: v;
};

However, I would like a way to not have to do that.

Is what I'm trying to do possible?

Tomas Reimers · Accepted Answer · 2021-10-01 16:02:48Z

1

Ah, so we can do this if we split out the inference to an identity function:

type KeyValueSameDict<Keys extends string> = {
  [v in Keys]?: v;
};

function inferGeneric<T extends string>(t: KeyValueSameDict<T>) {
  return t;
}

const foo = inferGeneric({ foo: 'foo', bar: 'bar' } as const);

answered Oct 1, 2021 at 16:02

Tomas Reimers

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3 Comments

akuiper Over a year ago

I think this is better than my answer. It works for objects with more than one key

sno2 Over a year ago

Making the key/value pairs was not optional in the question @Tomas Reimers.

Tomas Reimers Over a year ago

@sno2 apologies if I wasn't clear - what do you mean it wasn't possible for it to be optional?

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