typescript type predicate on array

Question

I am using a function that checks if a variable is an object and not null with:

function isRecord(input: any): input is Record<string, any> {
  return input !== null && typeof input === 'object';
}

The type predicate ist necessary, so typescript accepts:

if (isRecord(x)) {
  console.log(x["att"]);
}

I wrote another function that takes an array, but typescript complains with "Object is possibly 'null'":

function areRecords(list: any[]): list is Record<string, any>[] {
  return list.every(element => isRecord(element));
}

if (areRecords(x, y)) {
  console.log(x["att"]);
  console.log(y["att"]);
}

The same if I omit "is"

function areRecords2(list: any[]): boolean {
  return list.every(element => isRecord(element));
}

if (areRecords2([x, y])) {
  console.log(x["att"]);
  console.log(y["att"]);
}

Or if I use rest parameters:

function areRecords3(...list: any[]): boolean {
  return list.every(element => isRecord(element));
}

if (areRecords3(x, y)) {
  console.log(x["att"]);
  console.log(y["att"]);
}

How to do it right?

Silvermind · Accepted Answer · 2022-12-16 12:22:54Z

1

Using the array function, the signature provides that the array is a record holder and not x and y itself.

For example:

const list = [x, y];
if (areRecords(list)) {
    // Now typescript knows that list is Record<string, any>[];
    // However, it does not know what x and y is, because
    // the function signature only refers to the array.
}

What you can do is:

const list = [x, y];
if (areRecords(list)) {
    list.forEach(item => console.log(item["att"]));
}

or:

const attributes = [x, y].filter(isRecord).map(value => value["att"]);

answered Dec 16, 2022 at 12:22

Silvermind

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