Dice Loop- the number 1 can't be rolled

[clairelee0322]

I can run my dice loop but the only problem is that the number 1 can't be rolled no matter how large the number can roll..

i tried to change then value of x, rrRolls - x

 
                  arrRolls[rrRolls-1]++;
 

but it gets another number that never rolled.

sorry my English is not that good..
please help me out,, thanks..

#include <iostream>
 
#include <cstdlib>
 
#include <time.h>
 
#include <iomanip>
 
using namespace std;
 
int main(int argc, char* argv)
 
{
 
int rollDice();
 
 int rrRolls;
 
            int numRolls;
 
            int arrRolls[12];
 
            int dieRoll;
 
            int dice1;
 
            int dice2;
 
            int total_dice;
 
            int dieRoll1;
 
            srand((unsigned)time(0));
 
            cout << "How many times would you like to roll the dice?" << endl;
 
            cin >> numRolls;
 
            cout <<" "<< endl;
 
            for (int k = 0; k <12; k++)
 
            {
 
arrRolls[k] =0;
 
            }
 
            for (int i = 0; i <= numRolls; i++)
 
            {
 
                  dice1 =rand()%6+1;
 
                  dice2 =rand()%6+1;
 
rrRolls = dice1 + dice2;
 
                  dieRoll = rollDice();
 
                  dieRoll1 = rollDice();
 
                  cout << "D1 " << dieRoll1<< setw(10) << "D2: " << dieRoll << setw(10) << "total: " << dieRoll1 + dieRoll << endl;
 
                  arrRolls[rrRolls-1]++;
 
            }
 
            cout <<" "<< endl;
 
            cout << "Dice Rolls Total" << endl;
 
            for(int c =0;c < 12; c++)
 
            {
 
                  cout << c + 1 << " = " << arrRolls[c] << endl;
 
            }
 
return 0;
 
}
 
int rollDice()
 
{
 
      int result;
 
      result = (rand() % 6) + 1;
 
      return result;
 
}
 

[Ganon11]

You are rolling two six sided dice - thus, the sum can be anywhere from 2-12 (1+1 to 6+6). You are not getting a 1 because it is impossible.