Recursion of for's - c++

Question

I'm tried to figure out how to do it for quite of time and its not working as intended; I'm writing a code where there is 1 to k numbers, I need to find all possible combination without repeats. e.g. for 3: 1, 2, 3, 12, 13.
Example for counting 4-digits numbers with 1, 2, 3, 4, 5.
int k = 5;
for (int p = 0; p < k; p++)
{
for (int i = p+1; i < k; i++)
{
for (int j = i + 1; j < k; j++)
{
for (int h = j + 1; h < k; h++)
{
cout << p + 1 << i + 1 << j + 1 << h + 1 << endl;
}
}
}
}
And there is example for 3-digits number with 1, 2, 3.
int k = 4
for (int p = 0; p < k; p++)
{
for (int i = p+1; i < k; i++)
{
for (int j = i + 1; j < k; j++)
{
cout << p + 1 << i + 1 << j + 1 << endl;
}
}
}
I think that to count n-digits possible position without repeat i need n for's.
And i don't know how to do it without recursion which don't work when i do it.
My goal to get recursion which will count and print possible positions for n-digits.

Answer 1

I did recursion to count possibility myself, but love you guys for all your help.
My recursion is
void col(int ilosc)
{
static int st;
for (int i = st++; i < k; i++)
{
if (ilosc > 1)
col(ilosc - 1);
else
sposob++;
}
}
where ilosc is digits number and sposob is count of possible positions numbers.
NOTE: sposob and k is global variables.

Answer 2

I am not sure whether recursion is the best choice here, but you could do it like this:
typedef std::vector<int> IV;
IV getFirst(int k){
IV res;
for (int i=0;i<k-1;i++){res.push_back(i+1);}
return res;
}
bool getNext(IV& numbers,int i){
if (i==-1){return false;} // end of recursion
if (numbers[i]>i+1){return getNext(numbers,i-1);}
numbers[i]++;
return true;
}
bool getNext(IV& numbers){ // start of recursion
return getNext(numbers,numbers.size()-1);
}
int main() {
IV numbers = getFirst(5);
for (int i=0;i<numbers.size();i++){std::cout << numbers[i];}
std::cout << std::endl;
while(getNext(numbers)){
for (int i=0;i<numbers.size();i++){std::cout << numbers[i];}
std::cout << std::endl;
}
}

Answer 3

I think this will get you pretty close. I have an occasional repeat here, but this should set you on the right path.
const int max_depth = 5; // How long your string is
const int max_digit = 3; // Max digit you are counting to
int *nums = new int [max_depth];
void recurse_count(int depth)
{
if (depth < max_depth)
{
for(int i = depth; i <= depth+1; i++)
{
nums[depth] = i;
recurse_count(i+1);
}
}
else
{
for (int j = 0; j < max_depth; j++)
cout<<nums[j]+1;
cout<<endl;
}
}
int main()
{
recurse_count(0);
return 0;
}

Answer 4

My approach (still too early in the evening probably, I had problems with it)
namespace detail
{
void recurse_hlp(int min, int max, std::vector<int> vals, std::function<void(const std::vector<int>&)> f, std::size_t ptr)
{
if (ptr == vals.size())
f(vals);
else
{
for (int i = min; i <= max; ++i)
{
vals[ptr] = i;
recurse_hlp(min, max, vals, f, ptr + 1);
}
}
}
}
void recurse(int min, int max, int count, std::function<void(const std::vector<int>&)> f)
{
std::vector<int> vals(count);
detail::recurse_hlp(min, max, vals, f, 0);
}
void print(const std::vector<int>& vals)
{
for (int v : vals)
std::cout << v << " ";
std::cout << std::endl;
}
int main()
{
recurse(0, 5, 3, &print);
}
recurse gets a function accepting std::vector<int>, which contains all numbers from min to max up to count places.