Extending Error gives wrong stack trace information

Question

This question is not the same as Extended Errors do not have message or stack trace, which deals with es6 extends, but I'm using es5.

Here's my code

test.js

function FooError(message) {
    this.stack = (new Error(message)).stack;
    console.log('this.stack', this.stack);
    console.log('new_error_stack', (new Error(message)).stack);
    console.log('after');
}
FooError.prototype = new Error;

throw new FooError("foo");

This gives the following output : with node version 6.4 (node test.js)

this.stack Error
    at Object.<anonymous> (/tmp/test.js:7:22)
    at Module._compile (module.js:556:32)
    at Object.Module._extensions..js (module.js:565:10)
    at Module.load (module.js:473:32)
    at tryModuleLoad (module.js:432:12)
    at Function.Module._load (module.js:424:3)
    at Module.runMain (module.js:590:10)
    at run (bootstrap_node.js:394:7)
    at startup (bootstrap_node.js:149:9)
    at bootstrap_node.js:509:3
new_error_stack Error: foo
    at Error.FooError (/tmp/test.js:4:33)
    at Object.<anonymous> (/tmp/test.js:9:7)
    at Module._compile (module.js:556:32)
    at Object.Module._extensions..js (module.js:565:10)
    at Module.load (module.js:473:32)
    at tryModuleLoad (module.js:432:12)
    at Function.Module._load (module.js:424:3)
    at Module.runMain (module.js:590:10)
    at run (bootstrap_node.js:394:7)
    at startup (bootstrap_node.js:149:9)
after

/tmp/test.js:9
throw new FooError("foo");
^
Error
    at Object.<anonymous> (/tmp/test.js:7:22)
    at Module._compile (module.js:556:32)
    at Object.Module._extensions..js (module.js:565:10)
    at Module.load (module.js:473:32)
    at tryModuleLoad (module.js:432:12)
    at Function.Module._load (module.js:424:3)
    at Module.runMain (module.js:590:10)
    at run (bootstrap_node.js:394:7)
    at startup (bootstrap_node.js:149:9)
    at bootstrap_node.js:509:3

as you can see, the 'this.stack' doesn't show the same thing at all than the 'new_error_stack', eg, the this.stack shows the stack as if the error was thrown when the require was called. Also, the thrown stack is also "incorrect".

I took the code from https://stackoverflow.com/a/5251506/1993501 , and since I want to use instanceof FooError, I use the line FooError.prototype = new Error.

As you can see, the 'new_error_stack' gives the expected output, but I would have expected both values to be equal.

This has the effect that errors are not very useful for users of my library, see corresponding issue : https://github.com/open-xml-templating/docxtemplater/issues/245

What exactly is happening, and how can I prevent it and keep proper error types (eg being able to use instanceof) ?

Is there a reason you're not using parenthesis to surround the Error object declaration on these lines: `new Error(message).stack`? It strikes me as odd that it's not like this: `(new Error(message)).stack`, like it is in [this link](http://stackoverflow.com/questions/1382107/whats-a-good-way-to-extend-error-in-javascript/5251506#5251506) you gave. — Xetnus, Aug 27 '16 at 16:53
It wasn't intentional, but it seems to not have any influence, and I think both syntax are equivalent. I still edited my post so that the code doesn't differ. — edi9999, Aug 27 '16 at 17:13

score 1 · Answer 1 · edited May 23 '17 at 10:33

I found a possible solution to your problem, but I'm not entirely sure it fixes all of your issues. I'm using Error.prototype instead of new Error (which is what you had). I found this solution from the first answer provided to this question: How do I create a custom Error in JavaScript?

I also changed your console.log lines to concatenate the two strings, as I was using alert() to display the output for testing purposes.

Kevin Hakanson goes over why you may or may not want to go with this solution in his answer. Here's the code to answer your question:

function FooError(message) {
    this.stack = (new Error(message)).stack;
    console.log('this.stack ' + this.stack);
    console.log('new_error_stack ' + (new Error(message)).stack);
    console.log('after');
}

FooError.prototype = Error.prototype;

throw new FooError("foo");

You may want to use a try/catch block with that last line.

Ok, this seems to work, thanks. Do you have any idea of why it behaves differently? — edi9999, Aug 27 '16 at 18:03
I can't say I do; it's pretty strange that it behaves differently. If you ever run into issues using `Error.prototype`, however, check out the second answer on that link I gave you. — Xetnus, Aug 27 '16 at 18:14

Extending Error gives wrong stack trace information

1 Answers1