Attempting to access 2d array as a pointer - c++

Question

I am currently attempting to iterate through a 2d array to a function using pointer notation as an exercise. I found an example of how to do this on these forums; it's the if statement within the displayTable function. My compiler is giving me errors about the function call itself stating that the if statement from the displayTable function must be of a pointer type. Why is this not working? The example was up voted.
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
const int numRow = 6;
const int numCol = 26;
char letters[numRow][numCol] = {
{ 'm', 'w', 'r', 'u', 't', 'v', 'n', 'j', 'd', 'j', 'y', 'k', 'k', 'g', 'g', 'd', 'c', 'v', 'n', 'x', 'm', 'd', 'q', 'y', 'u', 't' },
{ 'y', 'e', 'r', 'y', 'e', 't', 'w', 'y', 'u', 'w', 'r', 's', 'f', 'h', 's', 'g', 'a', 'a', 'g', 'd', 'b', 'b', 'b', 'g', 'x', 'z' },
{ 'j', 'd', 'j', 'y', 'k', 'k', 'g', 'g', 'd', 'c', 'v', 'n', 't', 'w', 'y', 'u', 'w', 'r', 's', 'f', 'h', 's', 'g', 'a', 'a', 'g' },
{ 'y', 'e', 't', 'w', 'y', 'u', 'w', 'r', 's', 'f', 'h', 'j', 'y', 'k', 'k', 'g', 'g', 'd', 'c', 'v', 'g', 'a', 'a', 'g', 'd', 'b' },
{ 'e', 'r', 'y', 'e', 't', 't', 'v', 'n', 'j', 'd', 'j', 'y', 'k', 'w', 'r', 's', 'f', 'h', 's', 'g', 'g', 'g', 'd', 'c', 'v', 'g' },
{ 'y', 'u', 'w', 'r', 's', 'f', 'h', 's', 'g', 's', 'f', 'h', 's', 'g', 'a', 'a', 'g', 'd', 'w', 'y', 'u', 'w', 'r', 's', 'f', 'h' }
};
void displayTable(char[][26]);
int main()
{
char *ltrptr = 0;
ltrptr = &letters[0][0];
const int arraySize = 6 * 26;
int answer = 0;
char * arr[6][26];
displayTable(letters);
getchar();
return 0;
}
void displayTable(char ans[][26]){
//pas 2d array, then point to it
cout << " The table as it stands: " << endl;
char * ans1;
ans1 = &ans[0][0];
for (char * iter = &ans1[0][0]; iter != &ans1[0][0] + 6 * 26; iter++){
cout << &ans1[0][0] << endl;
}
}

Answer 1

The problem is that your ans1 is a simple pointer-to-char and you're trying to dereference it twice. The correct way to do the loop would be
for (char * iter = ans1; iter != ans1 + 6 * 26; iter++){
cout << *iter << endl;
}