强制所有剩余参数在 TypeScript 中是可选的

就我所知，只需键入undefined[]作为其余参数似乎就能达到目的:

function thisIsAfunction<TArgs extends undefined[] = []>(callback: (firstParam: string, ...args: TArgs) => void) { 
// −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−^^^^^^^^^
    /* Do something */ 
}

// This is okay ✅
thisIsAfunction((type: string, name?: string, id?: number) => {});

// This should cause a type error ❌
// (is it possible to force all args to be optional?)
thisIsAfunction((type: string, name: string, id: number) => {});

^{Playground link}

对于问题的扩展版本，您可以使用mapped tuple(在本例中称为Optionalized):

type Optionalized<T> = { [K in keyof T]: T[K] | undefined };

type SelectCommand<TArgs extends unknown[]> = (filter: any, ...args: Optionalized<TArgs>) => void;

class DataSource<TSelectArgs extends unknown[] = []> {
    select: SelectCommand<TSelectArgs>;
    constructor(config: {select: SelectCommand<TSelectArgs>}) {
        this.select = config.select;
    }
}

const ds = new DataSource({
    select: (filter: any, id?: number, type?: string) => { /* Do something */ }
})

ds.select('Search', 1, 'type'); // This should work
ds.select('Search'); // And this should always work

^{Playground link}

正如@SanBen在 comments 中指出的那样，上例中的Optionalized类型也可以替换为TypeScrip的Partial实用类型.