Re: replacing a function with another one

[Stefan Monnier <monnier@iro.umontreal.ca>]

> That line doesn´t tell me anything, it only confuses me.

Probably because you don't understand (lambda ...) yet.

> (defun replacement-fn (original-fn-arg0 original-fn-arg1)
>   (foobar)
>   (original-fn original-fn-arg0 original-fn-arg1)
>   (barfoo))

> (callinstead original-fn replacement-fn)

That's pretty much exactly what advice-add does for :around, except that
the original-fn is provided as an additional argument, so you have to write:

   (defun replacement-fn (orig-fun original-fn-arg0 original-fn-arg1)
     (foobar)
     (funcall orig-fun original-fn-arg0 original-fn-arg1)
     (barfoo))
  
   (advice-add 'original-fn :around 'replacement-fn)

> That would be clear and simple, assuming that calling the original
> function from within a function that is defined to be called instead of
> it always automatically calls the original function.

That's one of the reasons why the original function is passed as an
extra argument.  This way you can choose to either call (original-fn ...)
which will go through the advice recursively, or to call (funcall
orig-fun ...) which will call the unadvised version.

Magically making `original-fn' detect when it's called from its own
advice and skip the advice in this case would be not only less general,
but also pretty ugly to implement and unreliable.

> You could also have `callbefore', `callafter', etc.  Maybe it´s even
> possible to implement this, using add-advice.

Yup, just use :after or :before (in which case your advice won't receive
the `orig-fun' argument).


        Stefan