Derivation of the general function parameter

To provide a solution to this problem, let's consider the example below where a function is used. The function returns the same type that it accepts as the first argument:

function sample<U>(argument: (details: U) => U) {
    return null;
}

However, an issue arises when attempting to use this function. Instead of returning a string, the value of data appears as unknown:

sample((data) => {
    return "example"
})

What steps can be taken to resolve this problem?

Answer №1

Your purpose seems a bit unclear, but I believe my example can shed some light on it:

function sampleFunction<T>(param: T, callback: (data: T) => T): T {
    return callback(param);
}

let mySample = sampleFunction("example", (data) => {return data + Date.now()});
console.log(mySample);

You've defined a function "sampleFunction" with a generic type callback inside, but you're not utilizing this callback. Without using it, it's hard to determine the specific type being used.

The information on "Generics" proved helpful for me: https://www.typescriptlang.org/docs/handbook/2/generics.html#generic-types)

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