multi_vector.cpp

#include <iostream>
#include <vector>
#include <list>
#include <deque>
#include <set>
#include <map>
#include <algorithm>
#include <iterator>
#include <numeric>

using namespace std;
//vector<vector<vector<int> > > vec;  //3 dimension vec

template<size_t dimcount, typename T, size_t vecSize >
class multiVector {
	private:
		std::vector< multiVector< dimcount - 1, T, vecSize > > vec;
	public:

		multiVector() : vec(vecSize) {
			multiVector<dimcount - 1, T, vecSize>(); 
		}


		const multiVector<dimcount - 1, T, vecSize>& operator[] (int m) const {
			return vec.at(m); 
		}

		multiVector<dimcount - 1, T, vecSize>& operator[] (int m) {
			return vec.at(m); 
		}
};

template<typename T,size_t vecSize>
class multiVector<1, T,vecSize> {
	private:
		std::vector<T> vec;
	public:

		multiVector() : vec(vecSize) {}


		T& operator[] (const int index) {
			return vec[index];
		}

		T& operator[] (int m) const {
			return vec[m];
		}

};
//The first version of multiVector is the recursive class and the second version is the base class. We store a private member vector vec which is initialised in the default constructor and the recursive call is made, this calls the multiVector class until dimcount is 1 and then the base class version is invoked. The operator[] is overridden in both classes and is recursive until the base class where it returns the element in the base class vector. The main method for the program is below as it is in the demo.

int main() {

	vector<vector<vector<int> > > vec;  //3 dim vec

	//Fill vectors
	for (int i = 0; i < 5; i++) {
		vector<vector<int>> vv;
		for (int j = 0; j < 10; j++) {
			vector<int> ve;
			for (int t = 0; t < 3; t++) {
				ve.push_back(t);
			}
			vv.push_back(ve);
		}
		vec.push_back(vv);
	}

	//Print contents
	for (int i = 0; i < 5; i++) {
		for (int j = 0; j < 10; j++) {
			for (int t = 0; t < 3; t++) {
				cout << " element index " << t << " is " << vec[i][j][t] << "\n";
			}
		}
	}

	cout << "\n\n";
	multiVector<4, int, 5> vec4;
	cout << "accessing vec4[1][2][3][4][4] = 7 \n";
	vec4[1][2][3][4] = 7;  
	cout << "reading vec4[1][2][3][4][4] is ";
	cout << vec4[1][2][3][4] << "\n";

	//Just to prove good for N dimensions 100*100=10,000 vectors !!!.
	multiVector<4, int, 100> bigVec;

	cout << "\n\n";

	return 0;
}