Simple Sum Types In TypeScript Without Common Fields

Discriminated unions are well known in TypeScript. Their only (?) downside is they need a common property usually named kind or tag, e.g.:

type RenderProps<T> =
| { kind: 'children',
    children: (x: T) => ReactElement | null;
  }
| { kind: 'render',
    render: (x: T) => ReactElement | null;
  }

Which makes it a tad too wordy to use:

const RPComponent = (props: RenderProps<number>) => {
    switch(props.kind) {
    case ('children'):
        return props.children(42);
    case ('render')
        return props.render(42);
    }
}

// elsewhere
<RPComponent kind="render" render={n => <i>{n}</i>} />

Now, I’m fine with JS in templates (if you still think of JSX as of templates, and why not), but that unnecessary kind prop gets my goat.

So here’s a more compact solution:

type RenderProps<T> =
| { children: (x: T) => ReactElement | null;
    render?: never;
  }
| { children?: never;
    render: (x: T) => ReactElement | null;
  }

const RPComponent = (props: RenderProps<number>) =>
    (props.children || props.render)(42);

// and
<RPComponent render={n => <i>{n}</i>} />

It’s still a sum type (you can neither omit both children and render nor provide them both), but now you don’t need no stinking kind anywhere.

Mind that for some reason it’s not enough to declare the union variants as { chidlren: SomeType, render: undefined }. At least for JSX, TypeScript will want you to still specify a prop and give it a value of undefined. But render?: undefined (or never, which, I think, better conveys your intention) does the trick.