TypeScript conditional types not compiling

Question

Why does the following typescript not compile without errors?

const conditionalTypedFunction =
  (inputArg: string | string []):
    typeof inputArg extends string ? string : string[] 
  => {
    return inputArg;
  }

Here is a link to the TypeScript Playground: https://www.typescriptlang.org/play?ssl=3&ssc=2&pln=1&pc=1#code/MYewdgzgLgBKYBMCWUngIYBsAqBPADgKYIBiArmMKuDALwwAUSY+ZUAggE4DmAXDNE7NuMAD4CoQsCIDaAXQCU-KAUIgAZjGasOPGIQAeUQoggSpIgPznhMfoOHy6APhgBvALAAoGL5idCKDJOMC0WNi5uAG5vAF9vIA

I know that function overloads are an option. But I don't want to use them here.

cheesyMan · Accepted Answer · 2022-03-18 09:10:16Z

1

As inputArg is an Union Type string | string [], the condition typeof inputArg extends string always yields false. So, to TypeScript the function is always supposed to return a string[].

## EDIT / ADD ##

I added a link to this playground in the comments

edited Mar 18, 2022 at 9:10

answered Mar 17, 2022 at 16:31

cheesyMan

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7 Comments

philkunz Over a year ago

Ah, I see. ... Is there a way to solve this then without function overloads? I want the inputArg to be only of type string or string[] and the output type to be the same as inputArg when using the function.

JJWesterkamp Over a year ago

@philkunz You can use a generic to 'carry' the input type to the return type: const myFunction = <T extends string | string[]>(inputArg: T): T { ... }. See working example

philkunz Over a year ago

Whats strange though, that this doesn't work then: typescriptlang.org/play?#code/…

cheesyMan Over a year ago

@philkunz typescriptlang.org/play?#code/…

philkunz Over a year ago

@cheesyMan See, type inference still doesn't work. The assigned variables still are of type string | string[], regardless of the input. typescriptlang.org/play?#code/…

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