Typescript conditional return type with default argument value

Question

I'm trying to make a function return a conditional type based on the argument value but with a default value for the argument :

function myFunc<T extends boolean>(myBoolean: T = true): T extends true 
       ? string 
       : number {
       return  myBoolean ? 'string' : 1
    }

this throws an error Type 'true' is not assignable to type 'T'. 'true' is assignable to the constraint of type 'T', but 'T' could be instantiated with a different subtype of constraint 'boolean'.

I don't understand this error, since T is a boolean how come I can't assign true to it ?

I tried another approach with function overloads :

function myFunc(myBool: true): string
function myFunc(myBool: false): number
function myFunc(myBool = true) {
       return  myBool ? 'string' : 1
    }

myFunc()

but now typescript won't let me call myFunc() with no argument (even though it has a default value) and the first overload has an error This overload signature is not compatible with its implementation signature.

Is it even possible to do what I'm trying to achieve here in typescript and if son how ?

Answer 1

Your overload approach should work. You can make the parameter optional for the true overload:

function myFunc(myBool?: true): string
function myFunc(myBool: false): number
function myFunc(myBool = true): string | number {
    return  myBool ? 'string' : 1
    }

myFunc()