Print all arrangements in c++

Question

I wanna list all arrangement, the following is my sample code:

const unsigned char item1[] = {'b'};
const unsigned char item2[] = { 'A', 'C' ,'D'};
const unsigned char item3[] = {'1','2'};

int _tmain(int argc, _TCHAR* argv[])
{
    for (int i = 0; i < sizeof(item1) / sizeof(unsigned char); i++){
        for (int j = 0; j < sizeof(item2) / sizeof(unsigned char); j++){
            for (int k = 0; k < sizeof(item3) / sizeof(unsigned char); k++){
                printf("%c%c%c\n",item1[i],item2[j],item3[k]);
            }
        }
    }
    return 0;
}

This will print all arrangement, but I am worried about if the array item is from item1 to item99, the code is difficult to maintain. Is there a better solution to print all arrangement? Thanks!

Answer 1

You might store your "iterator" in vector, then you might do something like:

bool increase(const std::vector<std::string>& v, std::vector<std::size_t>& it)
{
    for (std::size_t i = 0, size = it.size(); i != size; ++i) {
        const std::size_t index = size - 1 - i;
        ++it[index];
        if (it[index] == v[index].size()) {
            it[index] = 0;
        } else {
            return true;
        }
    }
    return false;
}

void do_job(const std::vector<std::string>& v, std::vector<std::size_t>& it)
{
    for (std::size_t i = 0, size = v.size(); i != size; ++i) {
        std::cout << v[i][it[i]];
    }
    std::cout << std::endl;
}

void iterate(const std::vector<std::string>& v)
{
    std::vector<std::size_t> it(v.size(), 0);

    do {
        do_job(v, it);
    } while (increase(v, it));
}

Demo