type CheckFunction<T> = (value: T) => boolean;
type ResultFunction<T> = (arr: T[]) => T | undefined;
type FindFunction = <T>(checkCallback: CheckFunction<T>) => ResultFunction<T>;
const findElement: FindFunction = (checkCallback) => {
return (array) => {
for (let index = 0; index < array.length; index++) {
if (checkCallback(array[index])) {
return array[index];
}
}
return undefined;
};
};
const myNumbersArray = [1, 5, 4, 9];
const resultOne = findElement<number>((val) => val > 1)(myNumbersArray); // it works but explicitly defines the type 'number' in findElement<number>
const resultTwo = findElement((value: number) => value > 1)(myNumbersArray); // it works but explicitly defines the type 'number' in the check callback (value: number)
const resultThree = findElement((val) => val > 1)(myNumbersArray); // calling findElement() the way I want to, but both 'val' and 'resultThree' are 'unknown'
// ^
// Object is of type 'unknown'.
I'm deepening my understanding of Typescript and functional programming and encountered this issue:
I've created a higher order findElement function that should locate the first element in an array that meets a specific condition.
Here's my question:
Is there a way to enhance my typings so that the generic type T used in CheckFunction can be deduced from the type of elements in myNumbersArray without explicitly setting it as number? Additionally, the returned value from findElement()() should have the same type as the array elements or undefined if no element matches.
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