I want to declare a variable with two types via TypeScript, but the compiler gives me a type error.
interface IAnyPropObject {
[name: string]: any;
}
let a: IAnyPropObject | ((str: string) => any);
a.b = "bbbbbbbb"; /* type error */
a(""); /* type error */
I don't want to use any to declare.
I just want to constrain the variable to be only these types.
Because the code are so old and they are not TypeScript code.
Brief explanation of Mixin vs Union types.
Union: Either this or that type, but not both.
interface A { [name: string]: any; }
interface B { (str: string): string; }
type UnionType = A | B;
Mixin: A mix of this and that type at the same time.
interface A { [name: string]: any; }
interface B { (str: string): string; }
type MixinType = A & B;
Your code would work if you use a Mixin type, if that is your intention. Variable a can have a mix of both types at the same time.
If you use an OR type, that doesn't mean your object has two types at the same time, you have to test it and use the correct type with casting inside of the test.
See https://www.typescriptlang.org/docs/handbook/advanced-types.html
interface IAnyPropObject { [name: string]: any; }
type MyFunc = (str: string) => any;
let a: IAnyPropObject | ((str: string) => any);
if (a instanceof Function) {
(<MyFunc>a)("hi"); //passing a number will throw an error
} else {
(<IAnyPropObject>a).b = 'bbbbbbbb';
}
You could also create a custom type guard, it's explained in the documentation I linked to, then you would not have to cast it. There's a lot more to be said, I just scratched the surface since I'm answering from my phone, read the doc for all the details.
aso you cannot assign a property on it.