function findManyByShorthand(search)
{
    const found = [];

    for (const font of document.fonts) {
        if (conformsToShorthand(font, search)) {
            found.push(font);
        }
    }

    return found;
}
Returns an iterator of all `FontFace` from `document.fonts` that satisfy
the provided font shorthand.

@param {string} font

@returns {FontFace[]}