Convert string to char 2d array - c++

Question

I was wondering does anyone know how to convert a string into a 2d array? This was my attempt:
string w;
char s[9][9];
int p=0;
getline(cin, w);
while(p != w.size())
{
for (int k = 0; k < 9; k++)
{
for(int j = 0; j < 9; j++)
{
s[k][j] = w[p];
p++;
}
}
}
cout << "nums are: " << endl;
for(int k = 0; k < 9; k++)
{
for(int j = 0; j <9; j++)
{
cout << s[k][j];
}
}
But the numbers don't print out correctly. I want s[k][j] to print out everything in w but it simply prints out gibberish. I also noticed if i do string[81] then I get a whole bunch of errors. Could anyone help me? Thanks.

Answer 1

Try this:
const int NUM_ROWS = 9;
const int NUM_COLS = 9;
string w;
char s[NUM_ROWS][NUM_COLS];
getline(cin, w);
if (w.size() != (NUM_ROWS * NUM_COLS))
{
cerr << "Error! Size is " << w.size() << " rather than " << (NUM_ROWS * NUM_COLS) << endl;
exit(1);
}
for (int count = 0; count < w.size(); count++)
{
if (!isdigit(w[count]) && w[count] != '.')
{
cerr << "The character at " << count << " is not a number!" << endl;
}
}
for (int row = 0; row < NUM_ROWS; row++)
{
for(int col = 0; col < NUM_COLS; col++)
{
s[row][col] = w[col + (row * NUM_COLS)];
}
}
cout << "Nums are: " << endl;
for(int row = 0; row < NUM_ROWS; row++)
{
for(int col = 0; col < NUM_COLS; col++)
{
cout << s[row][col] << " ";
}
cout << endl;
}
Based on our chat, you might want this:
const int NUM_ROWS = 9;
const int NUM_COLS = 9;
string w;
char s[NUM_ROWS][NUM_COLS];
while (!cin.eof())
{
bool bad_input = false;
getline(cin, w);
if (w.size() != (NUM_ROWS * NUM_COLS))
{
cerr << "Error! Size is " << w.size() << " rather than " << (NUM_ROWS * NUM_COLS) << endl;
continue;
}
for (int count = 0; count < w.size(); count++)
{
if (!isdigit(w[count]) && w[count] != '.')
{
cerr << "The character at " << count << " is not a number!" << endl;
bad_input = true;
break;
}
}
if (bad_input)
continue;
for (int row = 0; row < NUM_ROWS; row++)
{
for(int col = 0; col < NUM_COLS; col++)
{
s[row][col] = w[col + (row * NUM_COLS)];
}
}
cout << "Nums are: " << endl;
for(int row = 0; row < NUM_ROWS; row++)
{
for(int col = 0; col < NUM_COLS; col++)
{
cout << s[row][col] << " ";
}
cout << endl;
}
}

Answer 2

You haven't described what you're trying to do very well and you haven't described the problem you're encountering, so the following is just based on guesses.
So it looks what you're trying to do is take a string like:
The quick brown fox jumped over the lazy dogs.
And put that into a 2D array like:
0 1 2 3 4 5 6 7 8
0 T h e q u i c k
1 b r o w n f o
2 x j u m p e d
3 o v e r t h e
4 l a z y d o g s
5 . x x x x x x x x
6 x x x x x x x x x
7 x x x x x x x x x
8 x x x x x x x x x
One thing wrong with your code is that as you copy values from w into s you don't ensure that the index p is actually within the bounds. You seem to have attempted to deal with this in the line that says while(p != w.size()); but that's an outer loop that does not protect p from being incremented out of bounds and used in the inner loops. Instead you'd have to put something like p++; if (p==w.size()) break; inside the inner most loop where you increment p. Or better yet, you should iterate over the string instead of over the array. Something like the following pseudo-code would replace your entire while(p){for(k){for(j){}}} set of loops.:
for(size_t i=0; i<w.size(); ++i) {
int k = compute target row from i
int j = compute target column from i
s[k][j] = w[i]
}
Also, here's some code to better visualize the array as you're debugging.
#include <iostream>
int main() {
char s[9][9] = {"The","quick","brown","fox","jumped","over","the","lazy","dogs."};
// your code to get input and copy it into the array goes here
//
// for(size_t i=0; i<w.size(); ++i) {
// int k = compute target row from i
// int j = compute target column from i
// s[k][j] = w[i]
// }
std::cout << " 0 1 2 3 4 5 6 7 8\n";
for (int i=0; i<9; ++i) {
std::cout << i;
for (int j=0; j<9; ++j)
std::cout << ' ' << s[i][j];
std::cout << '\n';
}
}
If you run this program without any changes the output should look like this:
0 1 2 3 4 5 6 7 8
0 T h e
1 q u i c k
2 b r o w n
3 f o x
4 j u m p e d
5 o v e r
6 t h e
7 l a z y
8 d o g s .