How do I print an 2D array by means of a pointer? c++

Question

I created a 2D array, and made a pointer to point to the first element. I'm trying to print the 2D array, using the pointer but I'm getting the following error. Not sure what I'm doing wrong here.

source_file.cpp: In function ‘int main()’: source_file.cpp:15:27: error: invalid types ‘char[int]’ for array subscript cout << pname[x][y] << endl;

                       ^

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

int main()
{
    char name[2][2] = {'E', 'f', 'g', 'r'};
  char* pname = &name[0][0];

  for (int x = 0; x<2; x++)
  {
    for (int y = 0; y<2; y++)
      {
        cout << pname[x][y] << endl;  
      }  
  }
}

Why do you want to take a pointer to `name`? That might help with a more substantial answer. — Tommy, Feb 04 '16 at 20:50
@ForeverStudent pname is supposed to be a pointer to my 2 D array. — rose, Feb 04 '16 at 20:50
Why not just use `name[x][y]`? What is the pointer usage supposed to do for you? — PaulMcKenzie, Feb 04 '16 at 20:51

Tommy · Answer 1 · 2016-02-05T00:57:11.957

5

pname is a pointer to char. So pname[x] is a char. Therefore pname[x][y] is an attempt to take the yth item from a char, just as if you'd typed char t; t[10];.

Assuming you want to pass a pointer because you want to pass by reference, you want to take an argument of char pname[][10].

If you just want to output the items linearly, use your current pname and step along for(int c = 0; c < 4; c++) cout << pname[c] << endl;.

edited Feb 05 '16 at 00:57

answered Feb 04 '16 at 20:51

Tommy

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@Tomyy oh I see.. Thank you! – rose Feb 04 '16 at 20:57

score 2 · Answer 2 · answered Feb 04 '16 at 20:50

2

What's happening here is, the char* type drops the array size information from the C-style 2-D array name. Because pname is just a char*, its users have no way of knowing the size and shape of the thing it's pointing to.

answered Feb 04 '16 at 20:50

comingstorm

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score 1 · Accepted Answer · answered Feb 04 '16 at 21:01

I would use something like:

  #define x_size 2
  [...]
  char (*pname)[x_size] = &name[0];

  for (int x = 0; x<2; x++)
  {
    for (int y = 0; y<2; y++)
      {
        cout << pname[x][y] << endl;  
      }  
  }

or:

  int x_size = 2;
  char* pname = &name[0][0];

  for (int x = 0; x<2; x++)
  {
    for (int y = 0; y<2; y++)
      {
        cout << pname[x*x_size + y] << endl;  
      }  
  }

score 0 · Answer 4 · edited May 23 '17 at 10:28

0

Do this:

int main() {
    char name[2][2] = {{'E', 'f'}, {'g', 'r'}};
    char pname[2][2];
    memcpy(pname, name, sizeof (char)*2*2);

    for (int x = 0; x<2; x++) {
        for (int y = 0; y<2; y++) {
            cout << pname[x][y] << endl;  
        }  
    }
}

It would output: E f g r with newlines inbetween

look up memcopy here

edited May 23 '17 at 10:28

Community

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answered Feb 04 '16 at 20:54

SegFault

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Answers the question by changing the parameters of the question and then answering the new question. Very Kobayashi Maru. – user4581301 Feb 04 '16 at 21:00

How do I print an 2D array by means of a pointer? c++

4 Answers4