Typescript: passing array type check result to a variable

Question

I have a piece of code where I need to know in a few different places whether or not a variable is an array (it can be an array or string). If it is, I map over the data. If not, I do something else.

Simplified code, which typescript complains about:

function myFunc(input: unknown[] | string) {

  const isArray = Array.isArray(input);
  
  if (isArray) {
    // this line throws an error, since `map` is not a valid method for strings
    return input.map(/* do stuff here*/);
  }

  return input;

}

If I change the line with the if statement to this:

  if (Array.isArray(input))

then typescript seems to understand that input is in fact an array.

I use that check in several places in the real code though, and would like to use a single variable for readability, rather than doing Array.isArray every time.

What's going on here?

Alex Wayne · Accepted Answer · 2021-07-06 18:58:49Z

4

Typescript 4.4+

What you want do does work in Typescript 4.4 (unreleased as of July 6, 2021)

See blog post detailing aliased conditions

Playground using nightly build

Typescript 4.3

In earlier versions of typescript your options are a bit more limited. The problem is that typescript is not smart enough to understand that const isArray is derived from Array.isArray(input). After that line, it's just a boolean value with no connection to any other value.

I would probably create an intermediate variable that only has a value if the input is an array. That way typescript can keep track of that type on one single variable.

Something like:

function myFunc(input: unknown[] | string) {
  const inputArray = Array.isArray(input) ? input : null

  if (inputArray) {
    return inputArray.map(str => str);
  }

  return input;
}

Playground

edited Jul 6, 2021 at 18:58

answered Jul 6, 2021 at 18:52

Alex Wayne

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2 Comments

Tushar Shahi Over a year ago

Your answer is innovative and works, but you have manipulated the variable to make it have either a falsy value or an array. From whatever I tried, any type check not inside the if-else body does not work. Can we say that with sureity?

Alex Wayne Over a year ago

Yes, it works, for sure. There is still a typecheck in the if. inputArray is type string[] | null. Then it's just a simple refinement. The if (inputArray) will filter out the null from that union, leaving you with just string[] in the conditional branch.

Tom HANAX · Accepted Answer · 2021-07-06 18:50:40Z

1

An Array.isArray is a special check that narrows the variable type. There is whole chapter on this: https://www.typescriptlang.org/docs/handbook/2/narrowing.html

Maybe you want to use pair of variables and use them later in the function.

const inputArray = Array.isArray(input) ? input : null;
const inputString = Array.isArray(input) ? null : input;

Now you can use corresponding variable if not null. But this only transforms one form of check to another form, so IMHO not so useful, really.

Of course, you can use one variable as any type, but you will loose all TS goodness.

const inputAny: any = input;

answered Jul 6, 2021 at 18:50

Tom HANAX

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1 Comment

Tushar Shahi Over a year ago

You could replace Array.isArray with const isArray = input instanceof Array; and it would still not work.

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