Initializing a 2D Array C++ - c++

Question

My goal is to read in a file at the command line, which is to be an adjacency matrix representation of a directed graph, and store the values in a 2D array. My plan is to read the number of integers on a line and use that as the size for my array(i.e. a 5 line input, with 5 integers per line would mean i would create a 5 x 5 array). However, to initialize an array, a constant value is needed, and counting the integers per line and storing it in a variable to use as my size parameter, does not allow me to create the array.
Sample input:
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
Code:
#include <iostream>
#include <sstream>
#include <fstream>
using namespace std;
int main(int argc, char **argv)
{
string currentLine;
int i, m = 0;
int count;
ifstream input(argv[1]);
int storage[10000];
printf("Original matrix: \n" );
if(input.is_open())
{
while(getline(input, currentLine))
{
istringstream iss(currentLine);
count = 0;
while(iss >> i)
{
if(iss.eof())//at end of each line, ends loop
{
count++;
storage[m] = i;
m++;
printf("%d \n", i);
break;
}
else
{
count++;
storage[m] = i;
m++;
printf("%d ", i);
}
}
}
}
int **matrix;
matrix = new int*[count];
for(int y = 0; y < count; y++)
matrix[y] = new int[count];
for(int r = 0; r < count; r++)
for(int c = 0; c < count; r++)
matrix[r][c] = storage[r+c];
printf("first = %d ", matrix[0][0]);
system("PAUSE");
return 0;
}
Based on my input, I should create a 5 x 5 array. But the line
int matrix[count][count];
Gives me an error saying that the "count" size parameter should be a constant. Is my way of counting the size of the input to use an invalid way of doing this, or is there a way to create a constant to use as my size param?

Answer 1

Instead of using 2D native C++ arrays, consider using a vector of vectors.
#include <vector>
#include <iostream>
int main() {
using namespace std;
int count = 5;
// create a matrix of 'count' rows by 0 columns
vector<vector<int>> matrix(count);
// resize the column count for each row
for (auto& r : matrix)
r.resize(count);
// use it just like an array
for (int i = 0; i < count; ++i)
matrix[i][i] = i;
for (int r = 0; r < count; ++r) {
for (int c = 0; c < count; ++c)
cout << matrix[r][c] << ' ';
cout << endl;
}
}

Answer 2

you will need to use a dynamic array
int *myArray; //Declare pointer to type of array
myArray = new int[x]; //use 'new' to create array of size x
myArray[3] = 10; //Use as normal (static) array
...
delete [] myArrray; //remeber to free memeory when finished.
http://www.cplusplus.com/forum/beginner/1601/
http://www.cplusplus.com/doc/tutorial/dynamic/

Answer 3

You are not going to get a 2-D array. You will have to work with Array of Array.
If you are willing to work with dynamic allocations, please do something like this:
int *myArray = new int [count]; // variable length array can get you into trouble here,
// if you are not careful as the inner dimension needs
// to be freed before the array goes out of scope.
for (/*iterate from 0 count*/) {
myArray[i] = new int [count];
}
myArray[m][n] to give you a semblance of what you wanted.