Hello,
I think it's not so much a question of 'what a shred is', as a more general issue of the scope of variables that are declared within a function (ie, they are not accessible outside the scope of said function). I'm about to leave my house, and not feeling quite clear enough to explain more completely, but I hope that's enough to point you in the right direction.
Cheers,
Peter
Dear list,
Shreds are objects (of type Shred) and functions aren't objects. This is all good and well but it gets harder when we spork a function and the function becomes a Shred and hence a Object. Consider this;
------------------------
fun void tester()
{
"boom" => string beep;
while(1)
{
<<<beep>>>;
second => now;
}
}
spork ~ tester() @=> Shred foo;
5::second => now;
"clik" => foo.beep; //this doesn't work
day=> now;
----------------------------------------
It's not so clear to me *why* the offending line doesn't work. Maybe I'm misunderstanding something or making a syntax error but I think that what we are seeing here is a loss of namespace that would be useful and I don't understand why this happens.
I also tried stuff along these lines;
----------------------
class Bar extends Shred
{
//stuff including variables and a constructor
}
------------------------------
If I do that I can instantiate a "Bar" but all it does is run it's constructor; I can't manage to spork a Bar or assign a sporked shred to a Bar. I can also make a array of Bar's but all that gets me is the constructor running;
---------------------------
class Bar extends Shred
{
static int number;
<<<number++>>>;
}
//print a series of numbers, no shreds in sight
Bar array[5];
----------------------------------
So, now I'm quite happy I said so many times that there are no "stupid n00b questions" because now it's my turn to ask exactly what a "Shred" is because I'm starting to feel like I have no idea. :¬)
Could somebody please explain to me exactly what's going on here?
Yours,
Kas.
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