Don’t use any as a type unless you are in the process of TypeScript has two special types, null and undefined, that have the values null and undefined respectively. However when void is a type of an argument it's not that clear what value can be used for it. Then, I figured out that TypeScript infers the type of the callback's parameter e to be an intersection(&) of MyMouseEvent and MyKeyboardEvent: e: MyEventObjects[K] >>>> e: MyMouseEvent & MyKeyboardEvent Generics. void exists in both JavaScript as an operator and in TypeScript as a primitive type. It is possible to specify void as a type parameter for generic function in TypeScript. No matter which expression is evaluated, void always returns undefined. The second part is the return type. We mentioned these briefly in the Basic Types section. As mentioned before, this is a required part of the function type, so if the function doesn’t return a value, you would use void instead of leaving it off. onSelect(someNumber: number) { } I've seen it both ways in different examples and wasn't sure if it is better to add the return type as void … Is there a way to specify a parameter for an argument of the void type? The void return type for functions can produce some unusual, but expected behavior. Don’t ever have a generic type which doesn’t use its type parameter. Nullable types. By default, the type checker considers null and undefined assignable to anything. It means the absence of any type. Allow me to quickly answer to "normal" use case of "How to define function overload types with TypeScript" with an example: I want a function that accepts a callback or returns a promise if none is provided: Important to note that void and any types overlap a bit, which may be weird at first. Contextual typing with a return type of void does not force functions to not return something. the result type of M is Void, or the result type of N is assignable to that of M. In this case, we're testing if => string is assignable to => void. We make it clear which is the return type by using an arrow (=>) between the parameters and the return type. any. S and T are object types and, for each member M in T, one of the following is true: M is a call, construct or index signature and S contains a call, construct or index signature N where. But, this didn't happen, because TypeScript didn't recognise the relationship between eventName: K and callback: (e: MyEventObjects[K]) => void. Solution When a Promise resolves with nothing, then you need to define this return type ( void ) in recent versions of TypeScript. onSelect(someNumber: number): void { } OR. void in JavaScript # void in JavaScript is an operator which evaluates the expression next to it. See more details in TypeScript FAQ page. Have your JS expressiveness and type it too. Then we get Type ‘1’ is not assignable to type ‘object’ and the code can’t be compiled with the TypeScript compiler and be run, since 1 is a primitive value. The void type is pretty much the opposite of the any type. Require explicit return types on functions and class methods (explicit-function-return-type) Explicit types for function return values makes it clear to any calling code what type is returned. And in both worlds void works a little bit different than most people are used to. And it works fine for return values of functions. In Typescript (using in an Angular project) for a method that returns nothing (void), which of the below is best practice? Did you forget to include ‘void’ in your type argument to ‘Promise’? Return type void. Instead of Object, use the non-primitive object type (added in TypeScript 2.2). 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