Read from stdin and store values into array C++ [duplicate]

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Possible Duplicate:

How to convert a 2d char array to a 2d int array?

I'm trying to read input from stdin and stop reading when encountering EOF. I need to store these values as integers in a 2x2 array, array[i][j].

I'm reading in 81 Sudoku characters (integers) from a single line + 2 more characters (\n and EOF) for a total of 83.

Ex:

STDIN -- > 123414292142341......2\n <EOF>

How do I only store t开发者_开发百科he numbers in each array[i][j]? and stop reading in when I encounter an <EOF> ?

So, after every 9 ints, I need to increment j to get the next row.

I'm looking to do this in C++

Thanks!

---

I have tried this so far

//Read in single line of 83 characters (81 sudoku integers + \n and //Store each integer into array of corresponding row and column

#include iostream

using namespace std;

string input_line;

int main()
{

  while(cin) {
    getline(cin, input_line);
  };


  return 0;
}

How do I tokenize the string "input_line" into a character of integers? and then to atoi to convert to int, and then finally store them in the Sudoku array??

---

OK, thanks almost done. but I now keep getting an invalid conversion from 'char' to 'const char*' error!

#include <iostream>
#include <stdlib.h>

using namespace std;

string input_line;

int main()
{

  char buffer[9][9];
  int sudoku[9][9];

  int v;

  cout << "Enter input: " << endl;

  cin >> (char*)buffer;

  for(int i = 0; i < 9; i++)
        for(int j = 0; j < 9; j++)
                v = atoi(buffer[i][j]);
                sudoku[i][j] = v;

---

If you want in C++ then you may choose to use C++ style cin. Following is pseudo code:

cout<<"Enter input:\n";
char sudoku[9][9];
cin>>&sudoku[0][0];

Here we are taking advantage of the fact that any digit will be less than < 256. So your table will be arranged inside the sudoku[][] automatically.