..() Purpose?

ID:157153
 
May 4 2010, 7:56 am
I was wondering, since I've seen this used in BYOND coding a couple of times at least, I was wondering what exactly this does in BYOND.
May 4 2010, 9:10 am
It calls the parent procedure, which can be thought of (only somewhat incorrectly) as "perform the default action". So, if you override Move(), a mob will no longer actually move when they try to move. However, when you call ..() in Move(), it will perform movement.

Note that in most cases, you actually want the line .=..(), which will also store the return value of the procedure in the variable . which is the default return value. In the case of Move(), for example, it returns 1 on success and 0 on failure, and if you simply call ..() and don't return anything yourself, it will no longer do this and can cause problems.
May 4 2010, 9:28 am
In response to Garthor
Ok, that answers that, but then how do you override something like Login()?
May 4 2010, 9:33 am
In response to CodeWeasel22
Simple, you just override it without adding "..()"

mob/Login()
  loc=locate("start")
  world<<"[src] logged in"
May 4 2010, 9:57 am
In response to Emasym
Technically, you're overriding it with or without the ..()
May 4 2010, 3:25 pm
In response to Garthor