this.callSuper() throws an exception

[Javier Peletier]

Hi!

 

I have found out this.callSuper() throws an uncontrolled exception if the base class does not implement the method.

 

While I think it makes sense that this call throws some exception if the base class does not implement the method, when coming to modules, well, the module implementer may be does not know if the base class the module will be included in will implement that method.

 

Therefore, I have modified core.js this way:

 

 

JS.Method.keyword('callSuper', function (method, env, receiver, args)

{

       var methods = env.lookup(method.name),

      stackIndex = methods.length - 1,

      params = JS.array(args);

 

       return function ()

       {

              var i = arguments.length;

              while (i--) params[i] = arguments[i];

 

              if (stackIndex==0)

                     return null;

 

              stackIndex -= 1;

              var returnValue = methods[stackIndex].apply(receiver, params);

              stackIndex += 1;

 

              return returnValue;

       };

});

 

This way, if someone attempts to use this.callSuper() when there is no function upwards, it just returns null and that’s it.

 

Anyway, the error thrown was cryptic, since it would try to call methods[-1].apply(), and the array would return undefined, thus crashing.

 

Cheers,

 

/j

 

[James Coglan]

On 18 January 2012 01:31, Javier Peletier <j...@peletier.com> wrote:

Hi!

 

I have found out this.callSuper() throws an uncontrolled exception if the base class does not implement the method.

 

While I think it makes sense that this call throws some exception if the base class does not implement the method, when coming to modules, well, the module implementer may be does not know if the base class the module will be included in will implement that method.

Ruby's approach to this is to make defined?(super) work so modules can check if they need to. It might be possible to remove the callSuper method when we reach the end of the stack, so you could do a similar thing in JavaScript. 

[j...@peletier.com]

I think I will stick to my modification. If I am always checking to see if can call a method, then I'd rather have the check inside the method and simplify my code. For me, callSuper means "do your stuff", either you call it before your code and then follow up, or you call it after to intercept, so I am not worried about it silently returning a null or 'undefined'.

Nevertheless, I think core.js should not throw an Index out of bounds exception, but rather a "no parent method to call" exception.

Cheers,
J

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From: James Coglan <jco...@gmail.com>

Sender: jsclas...@googlegroups.com

Date: Wed, 18 Jan 2012 10:10:39 +0000

To: <jsclas...@googlegroups.com>

ReplyTo: jsclas...@googlegroups.com

Subject: Re: this.callSuper() throws an exception

[James Coglan]

On 20 January 2012 09:32, <j...@peletier.com> wrote:

Nevertheless, I think core.js should not throw an Index out of bounds exception, but rather a "no parent method to call" exception.

I've implemented this on a branch: https://github.com/jcoglan/js.class/compare/3.0.x...safe-super

The effect of this is that callSuper is not defined if there is no parent method available. So in the following program:

var A = new JS.Class({

  foo: function() { this.callSuper() }

})

new A().foo()

The errors thrown by Node are:

3.0.x: TypeError: Cannot call method 'apply' of undefined

safe-super: TypeError: Property 'callSuper' of object #<:13529748362> is not a function