Optional Generic Type

As of TypeScript 2.3, you can use generic parameter defaults.

private logData<T, S = {}>(operation: string, responseData: T, requestData?: S) {
  // your implementation here
}

TS Update 2020: Giving void will make the generic type optional.

type SomeType<T = void> = OtherType<T>;

The answer above where a default value as the object is given make it optional but still give value to it.

Example with default type value is {}:

type BaseFunctionType<T1, T2> = (a:T1, b:T2) => void;

type FunctionType<T = {}> = BaseFunctionType<{name: string}, T>

const someFunction:FunctionType = (a) => {

}

someFunction({ name: "Siraj" });
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 
// Expected 2 arguments, but got 1.(2554)

Playground Link

Example with default generic type value is void

type BaseFunctionType<T1, T2> = (a:T1, b:T2) => void;

type FunctionType<T = void> = BaseFunctionType<{name: string}, T>

const someFunction:FunctionType = (a) => {

}

someFunction({ name: "Siraj" })

This is a good read on making generics optional.

Optional generic type in Typescript

Playground Link

If you're looking for an optional generic type within a Type/Interface declaration, this might help.

(came looking for this, only found answers dealing with generic function declarations. Siraj's answer got me on the right track.)

type ResponseWithMessage = {
  message: string;
};

interface ResponseWithData<T> extends ResponseWithMessage {
  data: T;
}

export type ResponseObject<T = void> = T extends void
  ? ResponseWithMessage
  : ResponseWithData<T>;

How about Partial?

Constructs a type with all properties of Type set to optional. This utility will return a type that represents all subsets of a given type.

You can write the overloading method like this:

private logData<T>(operation: string, responseData: T);
private logData<T, S>(operation: string, responseData: T, requestData?: S) {
    this.logger.log(operation + ' ' + this.url);
    if (requestData) {
        this.logger.log('SENT');
        this.logger.log(requestData);
    }
    this.logger.log('RECEIVED');
    this.logger.log(responseData);
    return responseData;
}

But I don't think you really need it, because you don't have to write this.logData<T, any>('GET', data) instead just write this.logData('GET', data). The T type will be infered

• void doesn't play well where some kind of object is expected. Edit: plays perfectly well, you just have to consider the fact that void will override anything you merge it with (void & string is basically void) - this is probably the one you want
• unknown
• {} basically the same as unknown

interface OffsetPagination {
    offset: number;
    limit: number;
}

interface PagePagination {
    page: number;
    size: number;
}

type HttpListParams<
    PaginationType extends OffsetPagination | PagePagination | void = void,
    Params = {
        filter?: string;
    },
> = PaginationType extends void ? Params : PaginationType & Params;

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