Optional parameters in Vanilla-JavaScript

Question

I don't know how to make function arguments to be optional in a certain case. The case is if I have at more than one argument and not the last one should be optional. I see the ejs templating engine solves this:

ejs.renderFile(templatesPath, options, (err, html) => {
    if (err) {
         console.log(err);   
         error = err;
    } else {
         res.write(html);
         res.end();
    }
})

So if I don't pass the options parameter, the callback function will still be recognized as a callback function and not as the options object - and the role, assigned to it, remains to be callback function and not the role of the missed parameter - as it would normally.

So how can it be solved - in JavaScript - that I still define 2-3 or more arguments (in the function definition) but if I don't pass an optional one - which is not the last - that it doesn't change the passed parameters role.

Answer 1

What matters is the order of the parameters. Therefore, you must repect it.

Just pass undefined, null or an empty object {} if the function does not support no value.

Below, options is declared but not valued (undefined). The order is respected, it will work.

You can keep this way: always declare variable to pass them to the function, and assign value to them only when you have it.

var options; 
ejs.renderFile(templatesPath, options, (err, html) => {
    if (err) {
         console.log(err);   
         error = err;
    } else {
         res.write(html);
         res.end();
    }
});