C++ Passing Dynamic Pointer to a 2D Array

Question

This is an extension to a previous question I previously asked. I can use a template to submit a dynamic 2D array and access the elements. Now, let's say I have a pointer to such an array. I am currently using the method shown here to assign my pointer to the original 2d array. I initially thought I could just change the expected input of the template function to (*A)[M][N] but that's a no-go. Am I missing some concept regarding pointers that someone could kindly explain?

header.h

#pragma once

#include <cstddef>
#include <iostream>
using namespace std;

template<size_t N, size_t M>
       void printPtr( int(*A)[M][N]) {
           for(int i=0; i < M; i++){
               for(int j=0; j<N; j++) {
                   cout << A[i][j] << " ";
               }
               cout << endl;
           }
       }

main.cpp

#include "header.h"
#include <iostream>
using namespace std;

int main() {

    int A[][3] = {
        {1,2,3},
        {4,5,6},
        {7,8,9},
        {10,11,12}
    };

    int (*ptrA)[3] = A;
    printPtr(ptrA);
}

Answer 1

_{If you're not interested in knowing why your code isn't working in the first place just skip to the end of this post}

The problem with your code is that you're "simulating" an array decaying by passing your pointer along:

int A[][2] = {
    {1,2,3},
    {4,5,6},
    {7,8,9},
    {10,11,12}
};

int (*ptrA)[3] = A;

This is confirmed by the following code which uses C++11 features:

#include <iostream>
#include <type_traits>
using namespace std;

template <typename T, typename U>
struct decay_equiv : 
    std::is_same<typename std::decay<T>::type, U>::type 
{};

int main() {

    int A[][4] = {
            { 1, 2, 3 },
            { 4, 5, 6 },
            { 7, 8, 9 },
            { 10, 11, 12 }
    };

    std::cout << std::boolalpha
              << decay_equiv<decltype(A), int(*)[3]>::value << '\n'; // Prints "true"
}

Example

You should either pass a non-decayed type (i.e. a type that has all the information regarding the array dimension) to the function via a pointer or a reference:

#include <cstddef>
#include <iostream>
using namespace std;

template<size_t N, size_t M>
void printPtr(int (*A)[M][N] /* Also a reference could work */) {
   for(int i=0; i < M; i++){
      for(int j=0; j<N; j++) {
         cout << (*A)[i][j] << " ";
      }
      cout << endl;
   }
}

int main() {

    int A[][6] = {
            { 1, 2, 3 },
            { 4, 5, 6 },
            { 7, 8, 9 },
            { 10, 11, 12 }
    };

    int(*ptrA)[4][7] = &A; // Not a decayed type
    printPtr(ptrA);
}

Example

Another solution would be to not use a pointer to the array in the first place or dereference it when passing a reference to the array:

template<size_t N, size_t M>
void printPtr( int(&A)[M][N]) {
     for(int i=0; i < M; i++){
         for(int j=0; j<N; j++) {
             cout << A[i][j] << " ";
         }
         cout << endl;
     }
}

...
printPtr(A);
printPtr(*ptrA);

Example

Answer 2

You defined the function parameter as a pointer to a two-dimensional array

int(*A)[M][N]) 

while are trying to call the function passing as the argument a pointer to a one-dimensional array

int (*ptrA)[3] = A;
printPtr(ptrA);

Moreover the function itself is invalid. It shall look like

template<size_t N, size_t M>
       void printPtr( int(*A)[M][N]) {
           for(int i=0; i < M; i++){
               for(int j=0; j<N; j++) {
                   cout << ( *A )[i][j] << " ";
               }
               cout << endl;
           }
       }

that is you have to use expression ( *A )[i][j] instead of A[i][j]

So you need to change the function the way shown above and to use appropriate pointer as the argument

int (*ptrA)[4][3] = &A;
printPtr(ptrA);

Of course it would be better to define the function parameter as reference to a two-dimensional array. For example

template<size_t N, size_t M>
       void printPtr( int(&A)[M][N]) {
           for(int i=0; i < M; i++){
               for(int j=0; j<N; j++) {
                   cout << A[i][j] << " ";
               }
               cout << endl;
           }
       }

In this case you could call the function like

printPtr( A );

Answer 3

It should be

template<size_t N, size_t M>
void printPtr(const int(&A)[M][N]) {
   for(int i=0; i < M; i++){
      for(int j=0; j<N; j++) {
         cout << A[i][j] << " ";
      }
      cout << endl;
   }
}

calling with:

printPtr(A);

or

template<size_t N, size_t M>
void printPtr(const int(*A)[M][N]) {
   for(int i=0; i < M; i++){
      for(int j=0; j<N; j++) {
         cout << (*A)[i][j] << " ";
      }
      cout << endl;
   }
}

calling with:

printPtr(&A);

BTW your pointer should be:

int (*ptrA)[4][3] = &A;