Creating N nested for-loops

Question

Is there a way to create for-loops of a form

for(int i = 0; i < 9; ++i) {
    for(int j = 0; j < 9; ++i) {
    //...
        for(int k = 0; k < 9; ++k) { //N-th loop

without knowing N at the compile time. Ideally I'm trying to figure out a way to loop through separate elements of a vector of digits to create each possible number if a certain amount of digits is replaced with different digits.

Answer 1

Here is a nice little class for a multi-index that can be iterated via a range-based for-loop:

#include<array>

template<int dim>
struct multi_index_t
{
    std::array<int, dim> size_array;
    template<typename ... Args>
    multi_index_t(Args&& ... args) : size_array(std::forward<Args>(args) ...) {}

    struct iterator
    {
        struct sentinel_t {};

        std::array<int, dim> index_array = {};
        std::array<int, dim> const& size_array;
        bool _end = false;

        iterator(std::array<int, dim> const& size_array) : size_array(size_array) {}

        auto& operator++()
        {
            for (int i = 0;i < dim;++i)
            {
                if (index_array[i] < size_array[i] - 1)
                {
                    ++index_array[i];
                    for (int j = 0;j < i;++j)
                    {
                        index_array[j] = 0;
                    }
                    return *this;
                }
            }
            _end = true;
            return *this;
        }
        auto& operator*()
        {
            return index_array;
        }
        bool operator!=(sentinel_t) const
        {
            return !_end;
        }
    };

    auto begin() const
    {
        return iterator{ size_array };
    }
    auto end() const
    {
        return typename iterator::sentinel_t{};
    }
};

template<typename ... index_t>
auto multi_index(index_t&& ... index)
{
    static constexpr int size = sizeof ... (index_t); 
    auto ar = std::array<int, size>{std::forward<index_t>(index) ...};
    return multi_index_t<size>(ar);
}

The basic idea is to use an array that holds a number of dim indices and then implement operator++ to increase these indices appropriately.

Use it as

for(auto m : multi_index(3,3,4))
{
    // now m[i] holds index of i-th loop
    // m[0] goes from 0 to 2
    // m[1] goes from 0 to 2
    // m[2] goes from 0 to 3
    std::cout<<m[0]<<" "<<m[1]<<" "<<m[2]<<std::endl;
}

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Answer 2

I use this solution:

unsigned int dim = 3;
unsigned int top = 5;
std::vector<unsigned int> m(dim, 0);
for (unsigned int i = 0; i < pow(top,dim); i++)
{
    // What you want to do comes here 
    //      |
    //      |
    //      v
    // -----------------------------------
    for (unsigned int j = 0; j < dim; j++)
    {
        std::cout << m[j] << ",";
    }
    std::cout << std::endl;
    // -----------------------------------

    // Increment m
    if (i == pow(top, dim) - 1) break;
    unsigned int index_to_increment = dim - 1;
    while(m[index_to_increment] == (top-1)) {
        m[index_to_increment] = 0;
        index_to_increment -= 1;
    }
    m[index_to_increment] += 1;
}

It can certainly optimized and adapted, but it works quite well and you don't need to pass parameters to a recursive function. With a separate function to increment the multi-index:

typedef std::vector<unsigned int> ivec;
void increment_multi_index(ivec &m, ivec const & upper_bounds)
{
    unsigned int dim = m.size();
    unsigned int i = dim - 1;
    while(m[i] == upper_bounds[i] - 1 && i>0) {
        m[i] = 0;
        i -= 1;
    }
    m[i] += 1;
}

int main() {

    unsigned int dim = 3;
    unsigned int top = 5;
    ivec m(dim, 0);
    ivec t(dim, top);
    for (unsigned int i = 0; i < pow(top,dim); i++)
    {
        // What you want to do comes here 
        //      |
        //      |
        //      v
        // -----------------------------------
        for (unsigned int j = 0; j < dim; j++)
        {
            std::cout << m[j] << ",";
        }
        std::cout << std::endl;
        // -----------------------------------

        // Increment m
        increment_multi_index(m, t);
    }

}

Answer 3

You could use a recursive function:

void loop_function(/*params*/,int N){
for(int i=0;i<9;++i){
    if(N>0) loop_function(/*new params*/,N-1);
}

This will call recursively to loop_function N times, while each function will iterate calling loop_function

It may be a bit harder to program this way, but it should do what you want

Answer 4

I wrote some C++ 11 code implementing a N-nested for-loop for myself. Here is the main part of the code that can be used as a single .hpp import (I named it nestedLoop.hpp):

#ifndef NESTEDLOOP_HPP
#define NESTEDLOOP_HPP
#include <vector>

namespace nestedLoop{

    class nestedLoop {
        public:
            //Variables
            std::vector<int> maxes;
            std::vector<int> idxes; //The last element is used for boundary control
            int N=0;
            int nestLevel=0;

            nestedLoop();
            nestedLoop(int,int);
            nestedLoop(int,std::vector<int>);

            void reset(int numberOfNests, int Max);
            void reset(int numberOfNests, std::vector<int> theMaxes);

            bool next();
            void jumpNest(int theNest);

        private:
            void clear();
    };

    //Initialisations
    nestedLoop::nestedLoop(){}

    nestedLoop::nestedLoop(int numberOfNests, int Max) {
        reset(numberOfNests, Max);
    }

    nestedLoop::nestedLoop(int numberOfNests, std::vector<int> theMaxes) {
        reset(numberOfNests,  theMaxes);
    }

    void nestedLoop::clear(){
        maxes.clear();
        idxes.clear();
        N = 0;
        nestLevel = 0;
    }

    //Reset the scene
    void nestedLoop::reset(int numberOfNests, int Max){
        std::vector<int> theMaxes;
        for(int i =0; i < numberOfNests; i++) theMaxes.push_back(Max);
        reset(numberOfNests, theMaxes);
    }

    void nestedLoop::reset(int numberOfNests, std::vector<int> theMaxes){
        clear();
        N = numberOfNests;

        maxes=theMaxes;

        idxes.push_back(-1);
        for(int i=1; i<N; i++) idxes.push_back(theMaxes[i]-1);
    }

    bool nestedLoop::next(){
        idxes[N-1]+=1;

        for(int i=N-1; i>=0; i--){
            if(idxes[i]>=maxes[i]) {
                idxes[i] = 0;

                if(i){ //actually, if i > 0 is needed
                    idxes[i-1] += 1;
                }else{
                    return false;
                }
            }else{
                nestLevel = i;
                break;
            }
        }
        return true;
    }

    void nestedLoop::jumpNest(int theNest){
        for(int i = N-1; i>theNest; i--) {
            idxes[i] = maxes[i]-1;
        }
    }
}
#endif // NESTEDLOOP_HPP

Here is an example with expected output:

#include <iostream>
#include "stlvecs.hpp"
#include "nestedLoop.hpp"

int main(){
    nestedLoop::nestedLoop looper;
    std::vector<int> maxes = {2, 3, 2, 2};
    looper.reset(4,maxes);
    int i = 0;
    while(looper.next()){
        std::cout << "Indices: " << looper.idxes << ", Last nest incremented: " << looper.nestLevel << std::endl;
        if(i == 5){
            std::cout << "...Jump Second Nest (index 1)..." << std::endl;
            looper.jumpNest(1);
        }
        i++;
    }
}

/* Expected output
Indices: 4  0 0 0 0 , Last nest incremented: 0
Indices: 4  0 0 0 1 , Last nest incremented: 3
Indices: 4  0 0 1 0 , Last nest incremented: 2
Indices: 4  0 0 1 1 , Last nest incremented: 3
Indices: 4  0 1 0 0 , Last nest incremented: 1
Indices: 4  0 1 0 1 , Last nest incremented: 3
...Jump Second Nest (index 1)...
Indices: 4  0 2 0 0 , Last nest incremented: 1
Indices: 4  0 2 0 1 , Last nest incremented: 3
Indices: 4  0 2 1 0 , Last nest incremented: 2
Indices: 4  0 2 1 1 , Last nest incremented: 3
Indices: 4  1 0 0 0 , Last nest incremented: 0
Indices: 4  1 0 0 1 , Last nest incremented: 3
Indices: 4  1 0 1 0 , Last nest incremented: 2
Indices: 4  1 0 1 1 , Last nest incremented: 3
Indices: 4  1 1 0 0 , Last nest incremented: 1
Indices: 4  1 1 0 1 , Last nest incremented: 3
Indices: 4  1 1 1 0 , Last nest incremented: 2
Indices: 4  1 1 1 1 , Last nest incremented: 3
Indices: 4  1 2 0 0 , Last nest incremented: 1
Indices: 4  1 2 0 1 , Last nest incremented: 3
Indices: 4  1 2 1 0 , Last nest incremented: 2
Indices: 4  1 2 1 1 , Last nest incremented: 3
*/

Answer 5

You may use recursion instead with a base condition -

void doRecursion(int baseCondition){

   if(baseCondition==0) return;

   //place your code here

   doRecursion(baseCondition-1);
}  

Now you don't need to provide the baseCondition value at compile time. You can provide it while calling the doRecursion() method.

Answer 6

I'm going to take the OP at face value on the example code that was given, and assume what's being asked for is a solution that counts through an arbitrary base-10 number. (I'm basing this on the comment "Ideally I'm trying to figure out a way to loop through seperate elements of a vector of digits to create each possible number".

This solution has a loop that counts through a vector of digits in base 10, and passes each successive value into a helper function (doThingWithNumber). For testing purposes I had this helper simply print out the number.

#include <iostream>

using namespace std;

void doThingWithNumber(const int* digits, int numDigits)
{
    int i;
    for (i = numDigits-1; i>=0; i--)
        cout << digits[i];
    cout << endl;
}

void loopOverAllNumbers(int numDigits)
{
    int* digits = new int [numDigits];
    int i;
    for (i = 0; i< numDigits; i++) 
        digits[i] = 0;

    int maxDigit = 0;
    while (maxDigit < numDigits) {
        doThingWithNumber(digits, numDigits);
        for (i = 0; i < numDigits; i++) {
            digits[i]++;
            if (digits[i] < 10)
                break;
            digits[i] = 0;
        }
        if (i > maxDigit)
            maxDigit = i;
    }
}

int main()
{
    loopOverAllNumbers(3);
    return 0;
}