now that I thought about it you're probably right and i don't need as
many as two bytes to show pheramones one should be sufficient
oky turns out I was doing all pheramone calculation before in bits :| i
don't know why but for some reason that is how i was doinng it, so now
i'll make the revised ones in bytes
now directional pheramones are only required for getting to resources
and back, atm the only resources we need are wheat,algea, wood and
stone(though wheat and stone are by far the most important) and to get
back from these locations to buildings.
vectorOfTeams<vectorOfTypes<pheramone>> directionalPheramoneVector
anyways if i do it in the following way
(12*2)*128^2+(2*4*9)*(128^2/2)
we have a result of 983 040
1.0 MB
the first half of that equation states that every square on the map will use 24 bytes for the two empty vectors
the second half showing that half the map would be covered in four
types of pheramones by two players (probably more then will
realistically occur)
now if we decide to have non-directional pheramones as well (to
indicate danger zones and length of time since last viewing) then we
can have that in the following way
vectorOfTeams<type[2]> nonDirectionalPheramones
so if it's initalized it will have both types put there
so
12*128^2+1*128^2
which is 212 992
so 0.2 MB
anyways well added together these make 1.2 MB.
approx 15 mb for a 512x512 map
anyways I NEED SOMEONE TO CHECK and see if my logic is correct, as
that's what i assume is going to happen in terms of ram usage,
especially with the vectors with directional pheramones
so if you have ANY questions please ask, btw _andrew said that
his tests on empty vectors took 12 bytes and that is what i used for my
calculations