Thread: Creating a multideminsional array with new

You might need to use a double pointer. You need to do something like this:

int **p;
p = new int*[n];
for(int i = 0; i < n; i++)
*(p+i) = new int[m];

Someone might be able to tell you another way though.

Then you can use p[x][y] = z;

Do you know any of the dimensions of the array when you define it? For example, if you didn't know the first dimension, but you knew that the second dimension was 2 you could define it as:

Code:

int (*array)[2];
array = new int[10][2];

If you don't know either dimension I think (don't quote me on this) that you can get away with:

Code:

int **array = new int*[10];
for (int i = 0; i < 10; i++)
{
	array[i] = new int[2];
}

Originally Posted by Link_26

You might need to use a double pointer. You need to do something like this:

...

Someone might be able to tell you another way though.

Then you can use p[x][y] = z;

Your code should be:

Code:

int **p;
p = new int*[n];
for(int i = 0; i < n; i++)
    p[i] = new int[m];

But you should avoid dynamic arrays in C++.

Originally Posted by Link_26

They both work. p[i] = *(p+i)

That's not all I fixed. You had p as a char **.

Subscripting is better, because it is easier to understand in a glance.

If you know in advance the number of rows, and how long each of the rows is, and you want all the allocated memory to be contiguous to reduce memory fragmentation and cache misses, you can allocate a single 1D array of the necessary length. If all of the rows are the same length, then do something like this (assuming m columns and n rows):

Code:

  int *p = new int[m*n];

Then you can access the (i, j) element with the notation p[m*i + j]. To clean up,

Code:

  delete[] p;

If the n rows have different lengths, then do

Code:

  int **p = new int*[n];
  p[0] = new int[num_elements];

where num_elements is the total number of elements in the array (equal to the sum of the lengths of all the rows). If the rows have length len0, len1, len2, ... then set

Code:

  p[1] = p[0] + len0;
  p[2] = p[1] + len1;
  // etc.

Then you can access the (i, j) element with the notation p[i][j]. To clean up,

Code:

  delete[] p[0];
  delete[] p;

Edit: If you don't know in advance either the number of rows, or the lengths of each of the rows, then you have to use vectors instead, to be able to do the dynamic resizing (in which case the memory isn't contiguous anymore).