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; +         


        

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

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;
}

Answer 4

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 5

You can use recursive call as:

void runNextNestedFor(std::vector<int> counters, int index)
{
     for(counters[index] = 0; counters[index] < 9; ++counters[index]) {
       // DO
       if(index!=N)
          runNextNestedFor(counters, index+1);
     }
}

Call it first time as:

std::vectors<int> counters(N);
runNextNestedFor(counters, 0);

Answer 6

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.