Is there a simpler way to do a one-line try without catch?

Question

When I do something like this:

let resolved = null;
try {
    resolved = require.resolve(modulePath)
} catch (e) {

}

I am wondering if there's a shorter syntax something like:

let resolved = null;
try resolved = require.resolve(modulePath)

Is there any way to forgive this line without opening up a catch block?

There's something like this but I'm looking for something more natural:

function t (fn, def) {
    let resolved = def;
    try {
        resolved = fn()
    } catch (e) {

    }
    return resolved;
}

Answer 1

No; it's intentional that JavaScript try-blocks must have either catch or finally. From https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/try...catch#Description:

The try statement consists of a try block, which contains one or more statements. {} must always be used, even for single statements. At least one catch clause, or a finally clause, must be present.

(And note that try { ... } finally { }, with no catch, does the opposite of what you want: you want everything to be caught and swallowed, whereas try { ... } finally { } doesn't catch anything.)

---

In your case, I think the best way to write it is:

let resolved;
try {
    resolved = require.resolve(modulePath);
} catch (e) {
    resolved = null;
}

which makes clear that resolved == null is the error-case.

(Better yet — add some logic to make sure that the exception you've caught is really the one you're expecting. You probably don't want to silently swallow exceptions that result from unintended bugs!)