I'm about to throw an exception using RangeError and wanted to check that I'm using it correctly and how to best catch it.

I have a function that could throw a RangeError OR a TypeError like this

function saveNumber(val) {
  // Only accept numbers.
  if (typeof val !== 'number') {
    throw new TypeError();
  }

  // Error if the number is outside of the range.
  if (val > max || val < min) {
    throw new RangeError();
  }

  db.save(val);
}

I'd like to call it and only deal with the RangeError. What's the best way to do this?

I'm about to throw an exception using RangeError and wanted to check that I'm using it correctly and how to best catch it.

I have a function that could throw a RangeError OR a TypeError like this

function saveNumber(val) {
  // Only accept numbers.
  if (typeof val !== 'number') {
    throw new TypeError();
  }

  // Error if the number is outside of the range.
  if (val > max || val < min) {
    throw new RangeError();
  }

  db.save(val);
}

I'd like to call it and only deal with the RangeError. What's the best way to do this?

• javascript
• try-catch

• Just don't throw a TypeError? – qwertynl Commented Dec 3, 2013 at 15:45
• another piece of code may want to call this function and deal with the TypeError – eddiec Commented Dec 3, 2013 at 15:45

3 Answers 3

try {
  saveNumber(...);
} catch (e) {
  if (e instanceof TypeError) {
    // ignore TypeError
  } 
  else if(e instanceof RangeError) {
    // handle RangeError
  }
  else {
    // something else
  } 
}

source

Straight from the MDN documentation on try - catch:

try {
    saveNumber(...);
} catch (e is instanceof RangeError) {
    // Do something
} catch (e) {
    // re-throw whatever you don't want to handle
    throw e;
}

slightly more elegant answer:

switch (error.constructor) {
    case NotAllowedError:
        return res.send(400);
    case NotFoundError:
        return res.send(404);
    default:
        return res.send(500);
}