As Matt McCutchen points this is a limitation of ReturnType
and in general conditional types and multiple overload signatures.
We can however construct a type that will return all overloaded return types for up to an arbitrary number of overloads:
function applyChanges1(input: string): number
function applyChanges1(input: number): string
function applyChanges1(input: number | string): number | string {
return typeof input === "number" ? input.toString() : input.length
}
function applyChanges2(input: number): string
function applyChanges2(input: string): number
function applyChanges2(input: number | string): number | string {
return typeof input === "number" ? input.toString() : input.length
}
type OverloadedReturnType<T> =
T extends { (...args: any[]) : infer R; (...args: any[]) : infer R; (...args: any[]) : infer R ; (...args: any[]) : infer R } ? R :
T extends { (...args: any[]) : infer R; (...args: any[]) : infer R; (...args: any[]) : infer R } ? R :
T extends { (...args: any[]) : infer R; (...args: any[]) : infer R } ? R :
T extends (...args: any[]) => infer R ? R : any
type RetO1 = OverloadedReturnType<typeof applyChanges1> // string | number
type RetO2 = OverloadedReturnType<typeof applyChanges2> // number | string
The version above will work for up to 4 overload signatures (whatever they may be) but can easily (if not prettily) be extended to more.
We can even get a union of possible argument types in the same way:
type OverloadedArguments<T> =
T extends { (...args: infer A1) : any; (...args: infer A2) : any; (...args: infer A3) : any ; (...args: infer A4) : any } ? A1|A2|A3|A4 :
T extends { (...args: infer A1) : any; (...args: infer A2) : any; (...args: infer A3) : any } ? A1|A2|A3 :
T extends { (...args: infer A1) : any; (...args: infer A2) : any } ? A1|A2 :
T extends (...args: infer A) => any ? A : any
type RetO1 = OverloadedArguments<typeof applyChanges1> // [string] & [number]
type RetO2 = OverloadedArguments<typeof applyChanges2> // [number] & [string]