Why the infered type of this function is `string | undefined`?

B

BenjaminOP•8/9/23, 11:45 AM

Even a simple version returns

string | undefined

string | undefined

:

export const myFunction = (value1?: string, value2?: string) => {
  if (value1 == undefined && value2 == undefined) {
    throw new Error("No value provided.");
  }

  return value1 != undefined ? value1 : value2;
};

export const myFunction = (value1?: string, value2?: string) => {
  if (value1 == undefined && value2 == undefined) {
    throw new Error("No value provided.");
  }

  return value1 != undefined ? value1 : value2;
};

export const myFunction = (value1?: string, value2?: string) => {
  if (value1 == undefined && value2 == undefined) {
    throw new Error("No value provided.");
  }

  return value1 != undefined ? value1 : value2;
};

export const myFunction = (value1?: string, value2?: string) => {
  if (value1 == undefined && value2 == undefined) {
    throw new Error("No value provided.");
  }

  return value1 != undefined ? value1 : value2;
};

A

Anna | DevMiner•8/9/23, 12:00 PM

this should be

  if (value1 == undefined || value2 == undefined) {
    throw new Error("No value provided.");
  }

  if (value1 == undefined || value2 == undefined) {
    throw new Error("No value provided.");
  }

  if (value1 == undefined || value2 == undefined) {
    throw new Error("No value provided.");
  }

  if (value1 == undefined || value2 == undefined) {
    throw new Error("No value provided.");
  }

A

Anna | DevMiner•8/9/23, 12:00 PM

then it'd be fine

B

BenjaminOP•8/9/23, 12:03 PM

Nope, you're changing the logic, I only want to throw if neither are defined (= both undefined). That's why TS should know at least one of them is defined and therefore the result is never

undefined

undefined

.

A

Anna | DevMiner•8/9/23, 12:05 PM

hmmm

B

BenjaminOP•8/9/23, 12:07 PM

I have to add

as string

as string

as string

as string and I hate that

D

dan•8/9/23, 12:10 PM

Its one of those cases where you cant get it to be inferred. The way it infers it technically is correct but I think as string is the only way.

A

Anna | DevMiner•8/9/23, 12:13 PM

how about

A

Anna | DevMiner•8/9/23, 12:14 PM

export const myFunction = (value1?: string, value2?: string) => {
  if (value1 !== undefined) return value1;
  if (value2 !== undefined) return value2;

  throw new Error("No value provided.");
};

export const myFunction = (value1?: string, value2?: string) => {
  if (value1 !== undefined) return value1;
  if (value2 !== undefined) return value2;

  throw new Error("No value provided.");
};

export const myFunction = (value1?: string, value2?: string) => {
  if (value1 !== undefined) return value1;
  if (value2 !== undefined) return value2;

  throw new Error("No value provided.");
};

export const myFunction = (value1?: string, value2?: string) => {
  if (value1 !== undefined) return value1;
  if (value2 !== undefined) return value2;

  throw new Error("No value provided.");
};

B

BenjaminOP•8/9/23, 12:31 PM

It was not exactly my use case, I am calling another function that takes a

string

string

as parameter. It works but it means I have to repeat the function call twice. Not ideal but maybe better than

as string

as string

as string

as string, I don't know. I guess that's the 2 solutions. ¯\_(ツ)_/¯

A

Anna | DevMiner•8/9/23, 12:33 PM

or you just use this as a separate function and then do something like this

export const actualFunction = (value1?: string, value2: string) => doSomeWork(myFunction(value1, value2));

export const actualFunction = (value1?: string, value2: string) => doSomeWork(myFunction(value1, value2));

export const actualFunction = (value1?: string, value2: string) => doSomeWork(myFunction(value1, value2));

export const actualFunction = (value1?: string, value2: string) => doSomeWork(myFunction(value1, value2));

B

BenjaminOP•8/9/23, 12:51 PM

yeah...

K

Kasper•8/9/23, 1:01 PM

export const myFunction = ({ value1 }: { value1?: string }, value2?: string) => {
  if (value1 && !value2) {
    return value1
  }

  if (!value1 && value2) {
    return value2
  }

  if (value1 && value2) {
    return value2
  }

  throw new Error("No value provided."); 
};

export const myFunction = ({ value1 }: { value1?: string }, value2?: string) => {
  if (value1 && !value2) {
    return value1
  }

  if (!value1 && value2) {
    return value2
  }

  if (value1 && value2) {
    return value2
  }

  throw new Error("No value provided."); 
};

export const myFunction = ({ value1 }: { value1?: string }, value2?: string) => {
  if (value1 && !value2) {
    return value1
  }

  if (!value1 && value2) {
    return value2
  }

  if (value1 && value2) {
    return value2
  }

  throw new Error("No value provided."); 
};

export const myFunction = ({ value1 }: { value1?: string }, value2?: string) => {
  if (value1 && !value2) {
    return value1
  }

  if (!value1 && value2) {
    return value2
  }

  if (value1 && value2) {
    return value2
  }

  throw new Error("No value provided."); 
};

K

Kasper•8/9/23, 1:02 PM

Not the best, but yeah

B

BenjaminOP•8/9/23, 1:02 PM

Yeah I think the

as string

as string

as string

as string is the most readable

K

Kasper•8/9/23, 1:03 PM

Yup

B

BenjaminOP•8/9/23, 1:03 PM

My post purpose was mainly to understand why TypeScript doesn't understand

D

dan•8/9/23, 1:04 PM

Because to the type system after the first if check, both are typed as string | undefined.

D

dan•8/9/23, 1:04 PM

and it cant pickup on the fact that if value1 is undefined, value2 has to be defined.

K

Kasper•8/9/23, 1:05 PM

Yeah, because it cant know which of them is not undefined.

B

BenjaminOP•8/9/23, 1:08 PM

but why if I only have one condition (

value !== undefined

value !== undefined

value !== undefined

value !== undefined), it manages, but not 2?

A

Anna | DevMiner•8/9/23, 1:09 PM

because then it's logical what changed

Why the infered type of this function is `string | undefined`?

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