Return either a wrapped or base type based on the condition

I am a beginner in TypeScript and I'm struggling to find the right combination of search terms to solve my issue. It seems like using a type condition could be helpful, but I still need to grasp how they function.

My goal is to pass a function that performs some sort of manipulation. If the function returns a specific type, then I would like to enclose it in a box; otherwise, return the original value.

For example:

class Box<T> {
    value: T;
    constructor(value: T) {
        this.value = value;
   }
}

function wrapIfString<T, TResult>(fn: (value: T) => TResult, value: T): 
   TResult extends string ? Box<string> : TResult {

   const result = fn(value);
   if (value instanceof String) {
       return new Box<string>(result);
   }

   return result;
}

However, this code does not compile. Can this be achieved in TypeScript?

Answer №1

Here is a solution for conditional return types that can sometimes lead to issues. Instead of using conditional return types, it may be cleaner and simpler to overload like this:

class Box<T> {
    value: T;
    constructor(value: T) {
        this.value = value;
   }
}


function wrapIfString<T>(fn: (value: T) => string, value: T): Box<string>
function wrapIfString<T, TResult>(fn: (value: T) => TResult, value: T): TResult
function wrapIfString<T, TResult>(fn: (value: T) => TResult, value: T): TResult | Box<string> {
   const result = fn(value);
    if (value instanceof String && typeof result === "string") {
       return new Box<string>(result);
   }
   return result;
}


const testType = wrapIfString((n) => "", "hello") // Box<string>
const testType1 = wrapIfString((n) => n * n, 5) // number

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