I am writing a game and I am having a challenge with my combat
function. All I want to do is find out how to group pieces that are in
the same space. There are two sides and all the units that are in the
same space fight. I want to add up the attack factors and defending
factors in the same space then figure out the odds so I can roll
against an odds table. Basically each piece holds its own x and y loc.
Here is what I have right now:
void fight(vector<un it>& at, vector<unit>& dt,
board * b , terrain * trn ){
vector<fightcla ssfighting;
/* adds attacking units to the fightclass vector if units are in the
same space they are to grouped in the same fight class elemet of
the
vector. */
for(int atu = 0; atu != at.size(); atu++){
if(fighting.siz e() == 0){
fightclass ft;
ft.addAu(at[atu]);
fighting.push_b ack(ft);
} else {
for(int lp = 0; lp != fighting.size() ; lp++){
if(at[atu].getXloc() != fighting[lp].getX() &&
at[atu].getYloc() != fighting[lp].getY()){
fightclass ft;
ft.addAu(at[atu]);
fighting.push_b ack(ft);
} else {
fighting[lp].addAu(at[atu]);
}
}
}
}
/* Adds defending units to the fightclass array. If x and y locs are
the same as attacking locations (are in the same space) they are
added to array for combat */
for(int dtu = 0; dtu != dt.size(); dtu++){
for(int lp = 0; lp != fighting.size() ; lp++){
if(dt[dtu].getXloc() == fighting[lp].getX() &&
dt[dtu].getYloc() == fighting[lp].getY()){
fighting[lp].addDu(dt[dtu]);
}
}
}
// Combat routine
for(int lp = 0; lp != fighting.size() ; lp++){ //handles combat
if(fighting[lp].canfight()){
int df = b->GetSpace(fight ing[lp].getX(), fighting[lp].getY());
float odds =
fighting[lp].getAtk()/fighting[lp].getDef();
//gets the defense bonus for the terrain in the space where
//combat takes place
int roll = rand() - trn[df].defend();
//get the die roll modified for terrain
odds = fighting[lp].getAtk() / fighting[lp].getDef();
//gets the attack to defence ratio.
if(odds < .5){
MessageBox(NULL , "Fighting! 1:3", "Info!", MB_OK);
return;
}
if(odds < 1){
MessageBox(NULL , "Fighting! 1:2", "Info!", MB_OK);
return;
}
if(odds < 2){
MessageBox(NULL , "Fighting! 1:1", "Info!", MB_OK);
return;
}
if(odds < 3){
MessageBox(NULL , "Fighting! 2:1", "Info!", MB_OK);
return;
}
if(odds < 4){
MessageBox(NULL , "Fighting! 3:1", "Info!", MB_OK);
return;
}
if(odds < 5){
MessageBox(NULL , "Fighting! 4:1", "Info!", MB_OK);
return;
}
if(odds < 6){
MessageBox(NULL , "Fighting! 5:1", "Info!", MB_OK);
return;
}
}
}
for(int lp = 0; lp != fighting.size() ; lp++){
fighting[lp].done();
}
fighting.clear( );
}
class fightclass{
/* Fightclass holds two arrays of units. The two arrays represent
units
in the same space elgible for combat. Array at represents the
attacking
forces and dt represents the defending units */
int xl;
int yl;
bool sides2; /* indicates if there are units in both attack and
defend
if there are units in both attack and defend then
combat
is to take place between the opposing sides. */
vector<unit>at; //attack units
vector<unit>dt; //defending units
public:
fightclass();
void addAu(unit au); //adds an addacking unit
void addDu(unit du); //adds a defending unit
int getX(){return xl;} //return the x coord
int getY(){return yl;} //returns the y coord
float getAtk(); //returns the total attack factors for one space
float getDef(); //returns the total defending factors for one space
bool canfight(); //if there are both attacking and defending units
void done(); //clears vectores after the turn
};
fightclass::fig htclass(){
xl = 0;
yl = 0;
}
void fightclass::add Au(unit u){
at.push_back(u) ;
xl = u.getXloc();
yl = u.getYloc();
}
void fightclass::add Du(unit u){
dt.push_back(u) ;
}
bool fightclass::can fight(){
if(at.size() && dt.size()){
return true;
}else {return false;}
}
float fightclass::get Atk(){
float total;
for(int lp = 0; lp != at.size(); lp++){
total +=at[lp].getAttack();
}
return total;
}
I know the logic is not the best but it is the best I can do. This is
tricky and I am pretty confused. I would like to find some way of
making the logic steps easier. There may be some simple error that I
overlooked or it can be totally screwed up.
