Question

I made this program just out of interest and wanted to make it better. My problem is that I want to make a nested for-loop to carry out the iterations but I can’t get my head around it, I have tried many times but my head is melting. Any help would be greatly appreciated. Also for some reason on windows and openSuse (from what I have seen) the program prints out some random characters after the expected output, a solution to this would be a great bonus. Thanks !

Sorry I didn’t make it clearer, the point of the code is to be able to theoretically generate every combination of letters from AAAAAAAA to ZZZZZZZZ.

1) No it’s not homework

#include <iostream>
using namespace std;

int main()
{
    char pass [] = {'A','A','A','A','A','A','A','A'};
    while(pass[0] != '[')
    {
        pass[7]++;

        if(pass[7]=='[')
        {
            pass[6]++;
            pass[7] = 'A';
        }
        if(pass[6] == '[')
        {
            pass[6] = 'A';
            pass[5]++;
        }
        if(pass[5] == '[')
        {
            pass[5] = 'A';
            pass[4]++;
        }
        if(pass[4] == '[')
        {
            pass[4] = 'A';
            pass[3]++;
        }
        if(pass[3] == '[')
        {
            pass[3] = 'A';
            pass[2]++;
        }
        if(pass[2] == '[')
        {
            pass[2] = 'A';
            pass[1]++;
        }
        if(pass[1] == '[')
        {

            pass[1] = 'A';
            pass[0]++;
        }

        cout << pass << endl;
    }
    return 0;
}

Answer 1

Maybe like this:

const char char_first = 'A';
const char char_last = '[';
const unsigned int passlen = 8;

while (pass[0] != char_last)
{
  ++pass[passlen - 1];

  for (unsigned int i = passlen - 1; i != 0; --i)
  {
    if (pass[i] == char_last)
    {
        ++pass[i - 1]; // OK, i is always > 0
        pass[i] = char_first;
    }
  }
}

For printing, include <string> and say:

std::cout << std::string(pass, passlen) << std::endl;

I took the liberty of making a few of the magic numbers into constants. If you’re ever going to refactor this into a separate function, you’ll see the merit of this.