如何在不抛出的情况下抛出错误

Question

So recently I was asked a question :

Is there a way to throw an error without using throw in javaScript ?

As far as I knew about errors, there was only one way to throw an error in JavaScript and that was using throw statement in JavaScript like so :

 var myFunc = () => { // some code here throw 'Some error' // in a conditional // some more code } try { myFunc() }catch(e) { console.log(e) } 

And not knowing any other way I said No, there is no other way . But now I'm wondering whether I was right ?

So the question is whether or not you can throw a custom error in JavaScript without using throw

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Restrictions :

• Kindly no using eval , Function .
• Don't use throw in your code

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Additional :

If you can throw an error without using the word Error

Answer 1

Generators do have a throw method that is usually used to throw an exception into the generator function code (at the place of a yield expression, similar to next ), but if not caught it bubbles out of the call:

(function*(){})().throw(new Error("example"))

Of course, this is a hack and not good style, I have no idea what answer they expected. Especially the "no division by zero" requirement is sketchy, since division by zero does not throw exceptions in JS anyway.

Answer 2

If all you want to do is throw an error then just do an invalid operation. I've listed a few below. Run them on your browser console to see the errors (tested on chrome and firefox).

 var myFunc = () => { encodeURI('\?'); // URIError a = l; // ReferenceError null.f() // TypeError } try { myFunc() }catch(e) { console.log(e); } 

Answer 3

 function throwErrorWithoutThrow(msg) { // get sample error try { NaN(); } catch (typeError) { // aquire reference to Error let E = typeError.__proto__.__proto__.constructor; // prepare custom Error let error = E(msg); // throw custom Error return Promise.reject(error); } } throwErrorWithoutThrow("hi") /* catch the error, you have to use .catch as this is a promise that's the only drawback, you can't use try-catch to catch these errors */ .catch((e) => { console.log("Caught You!"); console.error(e); }); 

Answer 4

I think, if you want to use try...catch blocks, you will need to throw something to get to the catch part. You can use pretty much everything that fails with an error to do that (like mentioned in the other posts).

If at some point, you want to run-or-die without having to write the throw statement yourself, you can always do an assertion:

const myFunction = () => {
  try {
    assert.fail();
    console.log(`Did not work :)`);
  } catch (e) {
    console.log(`OK`);
  }
};

Of course I would rather try to assert something positive (more Zen), that I need in order to continue.

Answer 5

const myTypeErr = () => `This is a Custom Err Message... \nand it`()

try {
  myTypeErr()
} catch(e) {} // catched

myTypeErr() // Uncaught TypeError: "This is a Custom Err Message... 
              // and it" is not a function