TypeScript Generics：用联合定义类型交集

Question

I'm building a TypeScript library that leverages some interfaces from another library. I'm trying to define a type from the intersection of a generic type and an interface I don't control, combined with a union between void , which has special meaning in the dependency library.

I've tried to create a minimal representation of the issue I'm facing.

export type AllProps<Props> = (Props & IDependecyProps) | void;

interface MyProps {
  disableCache: boolean;
}

function doTheThing(props: AllProps<MyProps>) {
  // Property 'disableCache' does not exist on type 'AllProps'.
  //  Property 'disableCache' does not exist on type 'void'.ts(2339)
  console.log(props.disableCache);
}

My goal is that the AllProps should allow you to specify either disableCache , and any properties in IDependecyProps , OR the type results in void . The library I depend on has a special meaning for void type, which makes it useful.

EDIT: I made the code sample too simple, forgot to add the generic type.

Answer 1

You can use a type assertion on your props and check for the existence of the property, since your example is boolean we need to check if it is not undefined

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interface IDependecyProps {
    something: number
}
export type AllProps = (MyProps & IDependecyProps) | void;

interface MyProps {
  disableCache: boolean;
}

function doTheThing(props: AllProps) {

  if ( typeof((props as MyProps).disableCache)!=='undefined' )
  console.log((props as MyProps).disableCache);
}
doTheThing({ disableCache: false, something:1})

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Answer 2

Simple if will refine the type and exclude the void :

function doTheThing(props: AllProps) {
  if (props) {
    console.log(props.disableCache); // props is narrowed to MyProps & IDependecyProps
  }
}

Control flow analysis narrows type of props inside if to MyProps & IDependecyProps .

Also if is needed anyway to prevent error at runtime ( props parameter can be undefined according to its type definition).

Playground