CODEWITHSUNDEEP
typescript
Pedro A
Pedro A

Reputation: 4333

How to declare a function that only accepts a value of type T if IsSomething<T> is true?

I have a complicated generic type IsInteresting<T> that becomes either the type true or false depending on what T I give it:

type IsInteresting<T> = (/* ... */) ? true : false;

Now I want to create a function that receives as first argument an interesting value, i.e., a value whose type U would cause IsInteresting<U> to be true. How can I accomplish this?

const operateOnInterestingValue: WhatFunctionTypeDoIPutHere = value => {
  // ...
};

let a: T1; // Assume `IsInteresting<T1>` would give `true`
let b: T2; // Assume `IsInteresting<T2>` would give `false`

operateOnInterestingValue(a); // Must compile
operateOnInterestingValue(b); // Must not compile

Upvotes: 0

Views: 27

Answers (1)

known-as-bmf
known-as-bmf

Reputation: 1222

If you slightly change IsInteresting<T> this way:

type IsInteresting<T> = (/* ... */) ? T : never;

You can then type your function like this:

const operateOnInterestingValue = <T>(value: IsInteresting<T>) => {
  // ...
};

The compiler will complain if you try to call operateOnInterestingValue with a variable whose IsInteresting<type> would resolve to never.

Upvotes: 1

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