Is it possible to determine the generic type without having an instance of it?
For example:
doThing<T extends Foo | Bar>(someArg: string): T {
if (T extends Foo) return functionThatReturnsFoo(someArg);
else return functionThatReturnsBar(someArg);
}
Is this kind of functionality achievable?
As pointed out by @jcalz, the challenge lies in the fact that the type system is lost during the compilation of TS to JS.
To address this issue, there are a few potential solutions that can be explored:
doThing function as an argument, leading to a union type returnIn my scenario, I opted to create two separate functions with similar logic. Although this may seem less efficient, it improves readability and facilitates testing.
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