Array within Array in C++

Question

hope you can help me.

Let's say I have two matrix arrays:

int n1[3][3]={{2,7,3},{1,8,7},{5,6,2}};
int n2[3][3]={{4,5,8},{5,5,6},{2,6,2}};

Then I can call one of them using

cout << n1[0][0] << endl;

which will return 2 of course but I want to get something like

cout << n[1][0][0] << endl; // Returns 2
cout << n[2][0][0] << endl; // Returns 4

Can you let me know how can I get this done?

You've already used 2D array, then why don't you use 3D array? — ikh
– ikh, Commented Aug 31, 2014 at 13:08
I didn't know something like this exists. Not everyone has years of experience in programming and documentation for C++ is very hard to understand. Also as Nawaz said that raw Arrays are not the best solution. I used std::array in my code. — Matt
– Matt, Commented Sep 1, 2014 at 10:52
Yes. In C++11, std::array is better than raw array except a bit long code. However, if you decide to use that, I recommend you to change your all raw arrays in your code into std::array for consistency. — ikh
– ikh, Commented Sep 1, 2014 at 15:12

Sarfaraz Nawaz · Accepted Answer · 2014-08-31 13:37:15Z

Define n as:

int n[2][3][3]= {
                 {{2,7,3},{1,8,7},{5,6,2}},
                 {{4,5,8},{5,5,6},{2,6,2}}
               };

Note that raw arrays in C++ are not good enough. Better use std::array:

std::array<std::array<std::array<int,3>,3>, 2>  n = {
                 {{2,7,3},{1,8,7},{5,6,2}},
                 {{4,5,8},{5,5,6},{2,6,2}}
               };

Well, that looks ugly — use typedef (or alias) to simplify the syntax:

 template<typename T, std::size_t M, std::size_t N, std::size_t P>
 using array3d = std::array<std::array<std::array<T,P>,N>, M>;

Then use it as:

array3d<int,2,3,3>  n = {
                          {{2,7,3},{1,8,7},{5,6,2}},
                          {{4,5,8},{5,5,6},{2,6,2}}
                        };

Thats better.

You could generalize the alias as:

#include <array> 

template<typename T, std::size_t D, std::size_t ... Ds> 
struct make_multi_array 
  : make_multi_array<typename make_multi_array<T,Ds...>::type, D> {}; 

template<typename T, std::size_t D>
struct make_multi_array<T,D> { using type = std::array<T, D>;  };

template<typename T, std::size_t D, std::size_t  ... Ds> 
using multi_array = typename make_multi_array<T,D,Ds...>::type;

Then use it as (read the comments for better understanding):

 //same as std::array<int,10> 
 //similar to int x[10] 
 multi_array<int,10>   x;   

 //same as std::array<std::array<int,20>,10>
 //similar to int y[10][20] 
 multi_array<int,10,20> y;   

 //same as std::array<std::array<std::array<int,30>,20>,10>
 //similar to int z[10][20][30]
 multi_array<int,10,20,30> z;

Hope that helps.

djechlin · Accepted Answer · 2014-08-31 13:05:44Z

1

Then you need to create a three-dimensional array and access the second row, first column, and first depth-column.

answered Aug 31, 2014 at 13:05

djechlin

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Comments

robbannn · Accepted Answer · 2014-08-31 13:23:50Z

1

int foo[x][y][z] = {};

This makes a multidimensional array of size x*y*z with all elements initialized to 0.

edited Aug 31, 2014 at 13:23

answered Aug 31, 2014 at 13:05

robbannn

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