Can the parse function be typed to return a string by default?

[mirkonasato]

What pain point are you perceiving?.

I see that after change #3341 released in v14.0.0 the value returned by the parse function is correctly inferred to be a string in TypeScript, but only if you explicitly pass async: false

const html: string = marked.parse(md, { async: false });

However it's still a Promise<string> | string by default

const html: Promise<string> | string = marked.parse(md);

forcing me to cast the result, which is not ideal:

const html = marked.parse(md) as string;

Describe the solution you'd like

Is there any reason why parse cannot treat the case where no options are passed in the same way as when passing async: false? So users can write

const html: string = marked.parse(md);

It seems to me the overloadedParse type could be simplified to

    type overloadedParse = {
      (src: string, options: MarkedOptions & { async: true }): Promise<string>;
      (src: string, options?: MarkedOptions | undefined | null): string;
    };

given that TypeScript will pick the first matching overload.

Happy to put together a PR if this makes sense, but I first wanted to check if I'm missing something because I'm not familiar with the full codebase.

[UziTech]

If an extension sets { async: true } then marked.parse(md) will return a Promise<string>

import { marked } from 'marked';

marked.use({ async: true });

marked.parse(''); // will return Promise<string>

[mirkonasato]

Understood, thanks.