Generic types with point-free functions

Question

Supose I have next funcion:

function cast<T>(v: any): T {
    return v as T;
}

Now, I have an array what I want to cast using map:

const arr = [1, 3, 5];

// works
arr.map(value => cast<number>(value));

// but I want
arr.map(cast<number>);

Is it possible? Can I pass generic on point-free map?

Answer 1

Actually, the method map from the Array prototype is generic.

Array<T>.map<U>(callback: (value: T, index: number, array: T[]) => U): U[]

So, by using your cast function with any type argument (like using it as a pointer-free) returns unknown. Totally expected.

You can cast your array like so

const arr = [1, 3, 5];
const re = arr.map<string>(cast);

Here is an interesting post about this.

---

I was reading this post months later and another idea just flashed in my mind.
Another solution could be a function wrapper to pass the type argument to the cast function explicitly.

type Fn<T> = (element: any) => T;

const cast = <T>(element: any): T => {
    return element as T;
}

const caster = <T>() => {
    return cast as Fn<T>;
}

const dut: any[] = [1, 2, 3, 4, 5, 6];
const casted = dut.map(caster<number>());

// casted is of type number[]

Although the main solution is still the best, this approach can solve some situations where the target API does not offer any overload to infer the type.

Hope it helps.