I have created a typescript util that return true if the parameter is a function and false if not, but when I use it to discriminate the way that I use a variable, typescript doesn't recognize the check.
util:
export function isFunction<T>(value: T | undefined): boolean {
return typeof value === 'function';
}
usage
isFunction(group) ? group(el) : group
That's not quite how user-defined type guards work:
function isFunction(x: unknown): x is Function { // note the 'is'
return typeof x === 'function';
}
const a = Math.round(Math.random()) ? () => 5 : null;
const b = isFunction(a) ? a() : null; // no error;
A rare actually valid use-case for the Function interface.
Function but it felt too weird to use lol
interface SpecialFunction extends Function { (x: number) => string; [someSymbol]: "I am special"; } because I needed the extra property typed.What you want is this Type predicate. No need to use generics.
export function isFunction(value: unknown): value is ((...args: any[]) => any) {
return typeof value === 'function';
}
function isFunction<T>(value: T | undefined): boolean {
return typeof value === 'function';
}
let imFn = (val: string) => {
console.log('i am a', val)
}
let imNotFn = 'not function';
function execute(anything: any){
console.log(isFunction(anything))
isFunction(anything) ? anything('function') : console.log(anything);
}
execute(imFn);
execute(imNotFn);
isFunctionis not caught by compiler ?