Element implicitly has an 'any' type because expression of type '"length"' can't be used to index type '{}'

Question

 function getLength<T>(arg: T): number {
 if (!arg.hasOwnProperty("length")) {
   return 0;
 }
 return arg["length"];
}

Here I am not able to access the value through bracket notation, this works in javascript but typescript is giving me issues on this.

This is the error in terminal thats running ts-node-dev

Answer 1

Using type predicates

type HasLength = { length: number };

function hasLength(arg: any): arg is HasLength {
  return (arg as HasLength).length !== undefined;
}

function getLength<T>(arg: T): number {
  if (arg && hasLength(arg)) {
    return arg.length;
  } else {
    return 0;
  }
}

console.log(getLength(3)); //0
console.log(getLength("333")); //3
console.log(getLength({ "1": 1, "2": 2, "3": 3 })); //0
console.log(getLength({ "1": 1, "2": 2, "3": 3, length: 3 })); //3
console.log(getLength([1, 2, 3])); //3
console.log(getLength(true)); //0
console.log(getLength(undefined)); //0
console.log(getLength(null)); //0

The reason for using arg && hasLength(arg) instead of hasLength(arg) is that calling undefined.length or null.length will throw an error.

Answer 2

in your function you're using a generic type named T and by default, it's defined as any, and any does not have a length property so you should extend your type so it should be defined in your generic like the following:

function getLength<T extends { length?: number, [key: string]: any }>(arg: T): number {
    if (typeof arg.length === 'undefined') {
        return 0;
    }
    return arg["length"];
}
console.log(getLength([1, 3, 4])) // --> 3
console.log(getLength({ length: 4 })) // --> 4
console.log(getLength({ a: 1 })) // --> 0