🥇I-Console roguelike ku-C++

Isingeniso

"I-Linux akuyona eyemidlalo!" - ibinzana eliphelelwe yisikhathi: manje kunemidlalo eminingi emangalisayo eqondiswe ngqo lolu hlelo oluhle. Kodwa noma kunjalo, ngezinye izikhathi ufuna okuthile okukhethekile okungakufanela ... Futhi nganquma ukudala le nto ekhethekile.

Isisekelo

Ngeke ngibonise futhi ngitshele yonke ikhodi (ayithakazelisi kakhulu) - amaphuzu abalulekile kuphela.

1.Umlingisi

Wonke amapharamitha omlingiswa abhalwe lapha (impilo, izikhali, isipiliyoni, njll.) Okuthakazelisa umdwebo kanye nesiqondiso sokunyakaza (okungatholakali okwamanje).

int x = 5, y = 5;
    hp = 100,
    maxhp = 100,
    dm    = 20,
    armor = 0,
    xp    = 0,
    level = 0,
    diff  = 10, // сложность
    pos   = 0; // направление

bool reg = 0,
       Mdm = 0, // бонусы
       ght = 0;

string color; // цвет будет использован в качестве индикатора состаяния героя

void hero()  // здесь происходит перемещение героя на координаты (x ; y)
{
  cout << "e[u " << "e[0;0H"; // восстановление позиции курсора, затирание пробелом
  for (int i = 0; i <= x; i++)
    cout << RIGHT;              // макрос "e[1C"
  for (int i = 0; i <= y; i++)
    cout << DOWN;             // макрос "e[1B"
  cout << "e[s" << color << "╬"; // сохранение позиции курсора
}

2.Ukuphatha

Indlela yokuhambisa uhlamvu isivele icacile (x—++, y—++). Kodwa ukucubungula ikhibhodi kuthakazelisa kakhulu:

char key;
char getkey()
{
  system("stty raw");
  key = getchar();
  system("stty cooked");
  return key;
}

Okusele nje ukusetha "izinhlamvu zokulawula". Ungakwenza usebenzisa ukushintsha, kodwa ngiyakuzonda.

switch(...) case .. : ... ; break kangcono kanje

#define KEY if (key ==
#define I ){
#define J ;}else

void keys()
{
  getkey();
    KEY 'a' I x-- ; pos = 1 J
    KEY......
}

Nobuhle! Imisebenzi ye-Loop bese ugijima esikrinini! Kodwa ngandlela-thile kulukhuni ... Futhi ikhesa ikhanya, nezinhlamvu ... Sizoyilungisa!

//До цикла
  cout << "e[?25l"; //отключаем отображение курсора
  system("stty -echo"); //отключаем эхо-ввод
  system("xset r rate 120 10"); // изменяем задержку на более плавную
//После цикла
//-------Return_normal_system_settings--------
  cout << "e[00m";
  system("reset");
  system("xset r rate 200 20");

Hewu! Iphesenti elilodwa selilungile!

3.Izwe elisizungezile

Lapha senza amalungu afanayo ka-x, y izingcezu zomhlaba nezingcezu ngokwazo (char o[N]), okufanayo ezilo kanye namabhonasi.

Dala umsebenzi world(int objx[N] .... objy[N] ... obj[N], ... objcolor[N]) ngokufanisa ne hero(), kodwa ngamapharamitha kanye neluphu eyengeziwe yokukhipha uhlu... ukuze sijabule, sidweba kuphela emkhakheni wokubuka (vis) (if (ox[k] < vis && oy[k]....))

Manje sigcwalisa isikrini ngezinhlayiya zomhlaba sisebenzisa amagumbi alula futhi angenalutho ngokwenqubo, ngesikhathi esifanayo sifaka izitha nezinto, ngokungahleliwe okuphelele esingakhohlwa ngakho. srand(time(NULL));

//------------------GENERATION---------------
void rooms()
{
  for (int i = 0; i <= 50; i++)
  {
    px[i] = rand() % 115 + 2;
    py[i] = rand() % 34 + 2;
    pl[i] = rand() % 5 + 5;
    ph[i] = rand() % 5 +  5;

    if (px[i] + pl[i] > 117) px[i] = 50 - pl[i] / 2; else
    if (px[i] < 2)           px[i] = 50 - pl[i] / 2; else
    if (py[i] < 1)           py[i] = 15 - ph[i] / 2; else
    if (py[i] + ph[i] > 37)  py[i] = 15 - ph[i] / 2;

    for (int j = 0; j <= i; j++)
    {
      while (px[i] > px[j] && px[i] < px[j] + pl[j])
        (px[i]+pl[i]/2 >= 55) ? px[i]++ : px[i]-- ;

      while (py[i] > py[j] && py[i] < py[j] + ph[j])
        (py[i]+ph[i]/2 >= 18) ? py[i]++ : py[i]-- ;

      while (px[i]+pl[i] > px[j] && px[i]+pl[i] < px[j] + pl[j])
        (px[i]+pl[i]/2 >= 55) ? px[i]++ : px[i]-- ;

      while (py[i]+ph[i] > py[j] && py[i]+ph[i] < py[j] + ph[j])
        (py[i]+ph[i]/2 >= 18) ? py[i]++ : py[i]-- ;
    }

    for (int j = 0; j <= i; j++)
    {
      while (px[j] + pl[j] >= 116) px[j]-- ;
      while (px[j] < 2)            px[j]++ ;
      while (py[j] < 1)            py[j]++ ;
      while (py[j] + ph[j] >= 37)  py[j]-- ;
    }
    tx[i] = px[i]+10; ty[i] = py[i]-3;

    if (i <= diff)
    {
      ex[i]  = px[i];
      ey[i]  = py[i];
      while (ex[i] < 10){ ex[i]++ ; epos[i] = 3 ;}
      while (ey[i] < 10){ ey[i]++ ; epos[i] = 1 ;}
      e[i]   = evar[pl[i]];
      ecolor[i] = "e[00me[31m";

      edm[i] = edmvar[pl[i]];
      ehp[i] = ehpvar[pl[i]];
      exp[i] = expvar[pl[i]];
    }
    rect(px[i], py[i], pl[i], ph[i]);
  }
}

void corrs()
{
  int pc, px, py;
  for (int i = 0; i <= 4; i++)
  {
    if (i < 2){
      px = 3;
      py = rand() % 33 + 3;
      pc = 110;
      line(px, py, pc, true);
      line(px, py+1, pc, true);
    } else {
      px = rand() % 100 + 3;
      py = 3;
      pc = 33;
      line(px, py, pc, false);
      line(px+1, py, pc, false);
    }
  }
}

4.Ukusebenzelana

Manje sidinga ngandlela thile ukugwema ukudlula ezindongeni nasezilo futhi sithole amabhonasi ezintweni.

Izintandokazi zethu ezethu futhi #define

#define TOUCH if (x == ox[i] && y == oy[i] && pos ==
#define HIT   x == ex[i] && y == ey[i] && pos ==
 for (int i = 0; i <= n; i++)
  {
    if (i <= diff)
    {
     if (Mdm) ehp[i]-=2 ; // если бонус "массовый урон" включен
     epos[i] = 0;

     if (ex[i] < x+5 && ex[i] > x-5 &&
         ey[i] < y+5 && ey[i] > y-5  )
     {
       edel(i); // функция переписывающая предыдущее положение противника
       if (ex[i] < x I ex[i]++ ; epos[i] = 1 J
       if (ex[i] > x I ex[i]-- ; epos[i] = 2 J
       if (ey[i] < y I ey[i]++ ; epos[i] = 3 J
       if (ey[i] > y I ey[i]-- ; epos[i] = 4 ;}
     }
   for (int j = 0; j <= n; j++) // столкновение моба со стенками
       while (ex[i] == ox[j] && ey[i] == oy[j] || ex[i] == ex[j] && ey[i] == ey[j] && j != i)
       {
         if (epos[i] == 1) ex[i]-- ; else
         if (epos[i] == 2) ex[i]++ ; else
         if (epos[i] == 3) ey[i]-- ; else
         if (epos[i] == 4) ey[i]++ ;
       }

     if (x == ex[i] && y == ey[i]) //  "битва"
      {
       if (ehp[i] > 1)
       {
         ehp[i] -= dm;
         (edm[i] < armor) ?
         hp -= 0 :
         hp -= edm[i]-armor;
       } else {
         ex[i] = ey[i] = -1;
         xp += exp[i];
         ehp[i] = 12;
       }
     }
     if (!ght) // если не призрак проверять столкновение игрока с врагами
     {
       if (HIT 1) y++ ;else
       if (HIT 2) x-- ;else
       if (HIT 3) y-- ;else
       if (HIT 4) x++ ;
     }
    }
    if (!ght) // то же, но со стенами
    {
      TOUCH 1 I y++ J
      TOUCH 2 I x-- J
      TOUCH 3 I y-- J
      TOUCH 4 ) x++ ;
    }
  }

5.Imenyu

Simane sibonise imenyu, sibale izinto, futhi sisebenzise i-getkey() ukucubungula ukukhetha komdlali. Sibhala umudwa wesimo somlingiswa, sebenzisa imenyu yokukala, sibhala i-backstory, futhi sithola lokho engikubiza ngokuthi “I-Subsoil”.

isiphetho

Lena into efana nale. Ungakwazi ukuyidlala ngemva kokulanda, ukuqaqa nokugijima kanje:

$ sudo chmod +x Subsoil-1.0/Subsoil

$ Subsoil-1.0/Subsoil

, noma, ekugcineni uphefumulelwe, zibhalele uhambo oluthandayo. Ngiyakuxwayisa kusengaphambili: umdlalo wami awulula!

Izixhumanisi

Ukukhiqizwa kwenqubo, Umkhuthazi.

Source: www.habr.com

I-Console roguelike ku-C++