typescript: declare module exports?

Question

How can i declare that a ts file or files exports some methods?

For example there is a es6 file toyota.ts

export const turnLeft: number = (param1: string, param2: number) => {
  // do it...
}
export const turnRight: number = (param1: string, param2: number) => {
  // do it...
}

A second es6 file honda.ts

export const turnLeft: number = (param1: string, param2: number) => {
  // do it...
}
export const turnRight: number = (param1: string, param2: number) => {
  // do it...
}

How can i make a type file which describes both these ts files. i.e. declare that the file has two exports which are called turnLeft and turnRight.

So that if I have a third file nissan.ts and it only has turnLeft, then when I do tsc it will tell me the error?

I have tried to enforce the toyota.ts by making a toyota.d.ts file with declare in it. But when tsc -w runs it does not even reload when i change the toyota.d.ts. Having the toyota.d.ts did nothing. So i could not even get the first step of getting one of the js file's exports to be checked, let alone all 3 files checked using the same declaration.

Thanks for any help.

Answer 1

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