Make function param accepts only keys from variable object

Question

I have an object with defined type for value:

type Type = { [key: string]: ValueType }

const variable: Type = {
    key1: valueType,
    key2: valueType,
    key3: valueType,
}

And I have a function func, which I want to be accepting only string with values from variable's keys:

func('key1')     // OK
func('key2')     // OK
func('key3')     // OK
func('keyother') // Error
func(3)          // Error

And this is what I have done when making type for func:

type FuncType = (param: keyof typeof variable) => any
const func: FuncType = ...

But I can only achieve one:

• typing for variable's value

or

• typing for func's param accept only variable's key

Not both.

• If I'm typing for variable's value const variable: Type = {, param has string type and I can pass any string to func call, which is wrong
• If I'm not typing for variable's value const variable: Type = {, func now typing param correctly but it makes variable accept anything as value.

Another way I can think about is predefined Type with list of keys ([key1, key2, ...]). But I don't want to maintain two list of the same thing. How can I achieve both of them without doing this way.

Typescript playground for this problem, which has some comments to describe problem more clearfully.

Answer 1

You have to use helper function that does nothing just validates type.

const makeType = <T extends Type = Type>(obj: T) => obj

const variable = makeType({
    key1: 0,
    key2: 0,
    key3: 0,
})

Playground

UPDATE: TypeScript 4.9 will make this possible with satisfies

Announcing TypeScript 4.9 Beta

const variable = {
    key1: valueType,
    key2: valueType,
    key3: valueType,
} satisfies Record<string, ValueType>