Check if object is of type (type is defined as array) - TypeScript

Question

Let's say I have this

export type Hash = [ hashtype, hash ];

export type hashtype = -16 | -43 | 5 | 6;
export type hash = Buffer;

I want to write something that will check whether an object is a Hash

not implemented

isHash = (obj: any) => {
    return (obj is Hash) // pseudo code, to implement
}

So that I would have such a return:

isHash(5)                         => false    //  no hash
isHash([25, <Buffer ad 30>])      => false    //  25 is not in hashType

isHash([5, <Buffer ad 30>])       => true     //  valid

Nicholas Tower · Accepted Answer · 2021-12-25 07:07:59Z

2

There isn't a general-purpose way to check whether a type matches. For your specific case, I would do something like this:

const isHash = (obj: unknown): obj is Hash => {
  if (!Array.isArray(obj)) {
    return false;
  }
  if (obj.length !== 2) {
    return false;
  }
  return [-16, -43, 5, 6].includes(obj[0]) && Buffer.isBuffer(obj[1]);
}

answered Dec 25, 2021 at 7:07

Nicholas Tower

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4 Comments

jsejcksn Over a year ago

You can just directly return all of those conditions infixed by && if you invert the first two.

jsejcksn Over a year ago

You seem to have deleted your last comment in between my reading it and preparing this playground link to send in response to it: tsplay.dev/W4yMAw

Nicholas Tower Over a year ago

Yeah, i deleted it because it was just an opinion question, so it doesn't really matter. Thanks for the feedback!

Stefan Vrecic Over a year ago

Thanks Nicholas, I'll just stick to that for now then. Maybe if I get around to it and it's not too complex, I can write an auto-generator :D

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