What causes a property type to become unassignable when passed through const in TypeScript?

I have a somewhat basic question that I can't seem to figure out on my own, so I'm hoping to get some help with it.

type SweetAlertPosition =
    'top' | 'top-start' | 'top-end' | 'top-left' | 'top-right' |
    'center' | 'center-start' | 'center-end' | 'center-left' | 'center-right' |
    'bottom' | 'bottom-start' | 'bottom-end' | 'bottom-left' | 'bottom-right';

Here's a simplified version of the main interface:

interface SweetAlertOptions {
    ...,
    position?: SweetAlertPosition
}
class Swal {
    mixin(options: SweetAlertOptions)
}

When I explicitly pass options to the .mixin call like this: Swal.mixin({ position: 'bottom' }), there are no errors. But if I define options in a const object like this:

defaultOptions = {
   position: 'bottom'
}

Passing it now produces an error - Swal.mixin(defauiltOptions)

Types of property 'position' are incompatible.
Type 'string' is not assignable to type '"top" | "top-start" | "top-end" | "top-left" | "top-right" | "center" | "center-start" | "center-end" | "center-left" | "center-right" | "bottom" | "bottom-start" | "bottom-end" | "bottom-left" | "bottom-right" | undefined'.

I've learned that I need to cast the String type to const like this:

defaultOptions = {
    position: 'botton' as const | as SweetAlertPosition
}

But why does passing options via a const object (defaultOptions) make the position property become a String-type, whereas passing it directly does not? O_o

Answer №1

Declaring a const variable means its type is determined solely by the value assigned to it, not its usage (full type inference). When initializing a variable with an object literal, the property types are widened so that other strings or numbers can be assigned to them - unless narrowed with as const.

When passing an object literal as an argument, its type is inferred from the parameter types of the function being called.

In this situation, using as const may not be necessary but it is advisable to do:

const defaultOptions: SweetAlertOptions = {
   position: 'bottom'
};

Answer №2

The issue arises when defaultOptions is of type { position: string } by default, unless you make a specific declaration like this:

const defaultOptions: SweetAlertOptions = {
   position: 'bottom'
}

By explicitly stating that defaultOptions should be of type SweetAlertOptions, there won't be any errors when passing it into mixin.

When the value is passed in explicitly as shown in the example, the compiler can interpret the object based on the context of the call.

See an example in the TypeScript Playground

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