I'm struggling with this code snippet:
function test(): any[] | string {
return [1,2]
}
let obj: any[] = test();
When I try to run it in vscode, I get the following error message:
[ts] Type 'string | any[]' is not assignable to type 'any[]'.
Type 'string' is not assignable to type 'any[]'.
I realize I need to add some kind of guard, but I'm having trouble grasping the concept. Can anyone offer some guidance?
When the test() function returns a type broader than the declared any[] variable in TypeScript, the compiler cannot guarantee that the function will always return an array instance. This leads to a complaint from the compiler. To resolve this issue, you can utilize the typeguards mentioned earlier, like so:
function test(): any[] | string
{
return [1, 2];
}
let val = test();
if (val instanceof Array)
{//no error here
let obj: any[] = val;
}
By implementing type guards, Typescript has the ability to narrow down a union type effectively. Utilizing a narrowing condition is simple enough, but integrating a type guard function enhances reusability and flexibility:
function isAnyArray(value: any[] | string): value is any[] {
return value instanceof any[];
}
This allows you to carry out your assignment as usual and then operate within the scope provided by your type guard:
let obj: string | any[] = test();
if (isAnyArray(obj)) {
//Utilize obj here as if it were an any[]
}
Given that your union is utilized in multiple instances, consider defining a type for it and using it instead of the manual union type:
type StringOrAnyArray = string | any[];
//...
function isAnyArray(value: StringOrAnyArray): value is any[] {
return value instanceof any[];
}
//...
let obj: StringOrAnyArray = test();
if (isAnyArray(obj)) {
//Utilize obj here as if it were an any[]
}
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