Programming Forums - Dynamic array size

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titaniumdecoy

Sep 3rd, 2006 11:41 PM

Dynamic array size

I want to create a 2-dimensional array of size N by N, where N is an integer read from a file.

int N;

std::ifstream in("file.in");

in >> N;



int **array = new int[N][N]; // ?

Why doesn't this work? How can I make it work?

grumpy

Sep 4th, 2006 2:33 AM

The short answer is that you can't do it, at least not in one step, because operator new [] only allocates a 1-dimensional array. A two dimensional array is an array one one dimensional arrays (more generally, an n-dimensional array is an array of (n-1) dimensional arrays.

One way of doing this is;

    int **array = new (int *)[N];   // array is a dynamically allocated array of pointers to int

       for (int i = 0; i < N; ++i)

       array[i] = new int [N];    // and each pointer in array is now a dynamically allocated array of int

Cache

Sep 4th, 2006 5:36 AM

You could use std::vector as an alternative to built in arrays. All the memory management is done for you.

typedef int element;

typedef std::vector<element> row;

typedef std::vector<row> two_d;



int N = 10;



two_d collection(N, row(N, 0));        // NxN elements initialezed to 0.



collection.at(0).at(0);



collection[0][0];



two_d::const_iterator td_iter = collection.begin();

row::const_iterator iter = td_iter->begin();

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