Shorter way to define custom JavaScript exceptions?

Question

I'm defining custom JavaScript exceptions like the code below. Is this proper? Isn't there a shorter way?

function InvalidModuleError(moduleName) {
    TypeError.apply(this);
    this.message = "module '" + moduleName + "' doesn't export any definitions";
    this.name = 'InvalidModuleError';
};

InvalidModuleError.prototype = Object.create(TypeError.prototype);


function DuplicateModuleError(moduleName) {
    TypeError.apply(this);
    this.message = "module '" + moduleName + "' is already defined";
    this.name = 'DuplicateModuleError';
};

DuplicateModuleError.prototype = Object.create(TypeError.prototype);

Edit: Eventually, after Oriol's suggestion, I've made a function that generates exceptions and looks like this:

function makeException(parentObject, parentClass, name, message) {
    var shortName = name.split('.');
    shortName = shortName[shortName.length - 1];
    parentObject[shortName] = function() {
        this.message = (typeof message === 'function') ? message.apply(null, arguments) : message;
    };
    parentObject[shortName].prototype = Object.create(parentClass.prototype, {'name': {'value': name}});
}

Oriol · Accepted Answer · 2013-11-24 23:01:47Z

2

Your code can be simplified a bit:

function InvalidModuleError(moduleName) {
    this.message = "module '" + moduleName + "' doesn't export any definitions";
}
InvalidModuleError.prototype = new Error();
InvalidModuleError.prototype.name = 'InvalidModuleError';

function DuplicateModuleError(moduleName) {
    this.message = "module '" + moduleName + "' is already defined";
}
DuplicateModuleError.prototype = new Error();
DuplicateModuleError.prototype.name = 'DuplicateModuleError';

But if you want to have lots of custom exceptions, better use

function newModuleError(errorName, message) {
    var moduleError = function(moduleName) {
        this.message = message.replace('%s', moduleName);
    };
    moduleError.prototype = new Error();
    moduleError.prototype.name = errorName;
    return moduleError; 
};

var InvalidModuleError = newModuleError('InvalidModuleError', "module '%s' doesn't export any definitions"),
    DuplicateModuleError = newModuleError('DuplicateModuleError', "module '%s' is already defined");

To throw them:

throw new InvalidModuleError("moduleName1");
throw new DuplicateModuleError("moduleName2");

edited Nov 24, 2013 at 23:01

answered Nov 24, 2013 at 20:45

Oriol

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3 Comments

HankMoody Over a year ago

I'm defining exceptions. It seems, that without prototype inheritance (MyException.prototype = Object.create(BaseException.prototype);) exceptions are not printed in browser's console properly (maybe I'm doing something wrong).

Oriol Over a year ago

@PythonFanboy I have added a toString method to display it well.

Oriol Over a year ago

@PythonFanboy I have updated it to throw real errors instead of objects with .toString

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