Is it possible to make a function have either mandatory or optional parameters based on conditional types in TypeScript? This is what I've got so far: const foo = <T extends string | number&gt...

The following function was largely lifted from the typescript handbook section on using conditional types, yet it doesn't work: function test<T extends boolean>(a: T): T extends true ? strin...

I was wondering if I can have conditional types in TypeScript? Currently I have the following interface: interface ValidationResult { isValid: boolean; errorText?: string; } But I want to rem...

I am trying to understand how TypeScript conditional type works. Here is my code. There are type errors: interface MyType { name: string; } const testFunc = <T extends MyType | string>( w...

So i was going through the documentation on typescript and am not able to get my head around this concept. So the documentation states :- In instantiations of a distributive conditional type T exte...

I may have missed something in the docs, but I can't find any way in typescript to get the types of the parameters in a function. That is, I've got a function function test(a: string, b: num...