Comparing a string with an object in Typescript?

Question

function printAll(strs: string | string[] | null) {
  if (strs && typeof strs === "object") {
    for (const s of strs) {
      console.log(s);
    }
  } else if (typeof strs === "string") {
    console.log(strs);
  }
}

https://www.typescriptlang.org/docs/handbook/2/narrowing.html

Here I have broken the big if statement into two parts as follows. The ... show the part that I am not referring to.

I can understand that here if ( ... typeof strs === "object") { we are comparing the type of the variable because it can contain other types too.

I do not understand this: if (strs === "object" ... ) {. What is the purpose of this statement?

please re-read your question, seems like you have a typo and it's not clear what you are asking — thedude
– thedude, Commented Apr 26, 2021 at 9:24
if (strs && typeof strs === "object") is saying "if strs is truthy and strs's type descriptor is "object"". the double amerpsand (&&) delimits the two conditions. — user3282374
– user3282374, Commented Apr 26, 2021 at 9:28
Like mentioned, your question is slightly muddled. But I'm assuming your asking why the if (typeof strs === 'string'), and not just use } else {,.. It's because strs could still be null. — Keith
– Keith, Commented Apr 26, 2021 at 9:36

Felix Kling · Accepted Answer · 2021-04-26 09:40:54Z

2

if (strs && typeof strs === "object")

is not the same as

if (strs === "object" && typeof strs === "object")

which is what you seem to be thinking. It's equivalent to

if (Boolean(str) && (typeof strs === "object"))

Why is it necessary to check whether str is a "truthy" value? Because typeof returns "object" for two data types: Object values and Null values (i.e. null):

console.log(typeof {});
console.log(typeof null);

So if you only did if (typeof strs === "object"), strs could still be null and therefore the for loop would fail (or in this case TypeScript would complain).

The code could have been written as

if (strs !== null && typeof strs === "object")

to be more explicit.

answered Apr 26, 2021 at 9:40

Felix Kling

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2 Comments

user3282374 Over a year ago

strs !== null doesn't eliminate undefined

Felix Kling Over a year ago

@CallumMorrisson: True, but that should already be a compile time error because the type of strs is string | string[] | null.

thedude · Accepted Answer · 2021-04-26 09:41:30Z

1

I've re-ordered the code a little and added some comments. Hope it answers you question:

function printAll(strs: string | string[] | null) {
  // handle the null case
  if (strs == null) {
    return
  }

  // handle the string[] case
  if (typeof strs === "object") {
    for (const s of strs) {
      console.log(s);
    }
    return
  }

  // if we reached this point strs is a string, we could check here if (type strs === 'string') but it's redundant
  console.log(strs);
}

edited Apr 26, 2021 at 9:41

answered Apr 26, 2021 at 9:35

thedude

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3 Comments

Felix Kling Over a year ago

It's not equivalent to the original code though. The case where str === "" is different.

cdimitroulas Over a year ago

I would also recommend Array.isArray instead of typeof x === "object" for checking if something is an array

thedude Over a year ago

I wanted to stay close to the OP's question

Tony Tom · Accepted Answer · 2021-04-26 09:43:09Z

function printAll(strs: string | string[] | null) {
  if (strs && typeof strs === "object") {
    for (const s of strs) {
       console.log(s);
    }
  } else if (typeof strs === "string") {
    console.log(strs);
  } else {
    // do nothing
  }
}

In the printAll function, you can pass inputs as string, string[], or null

let fruits: string[] = ['Apple', 'Orange', 'Banana'];
printAll(fruits);
let fruit: string = 'peach';
printAll(fruit);

So in case of string array type of strs is object and in the other case its string as expected.

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