How do I go about creating a dynamic 3d array . . . int arr[][][3]; Then later i have . . . arr = new int [w][h][3]; This does not work, complains about zero sized array declarations. i have tried a multitude of other ways of defining it but none work. Anyone have any ideas thanks

Since you declared your variable as an array, you cannot assign a new address to it:

  int foo[3];...foo = new int[3];  

is forbidden.

The solution : declare your variable as a pointer (yep, can't live without those).

  int* foo;...foo = new int[3];  

If you want a column of them :

  int i;int **foo = new (int*)[h]for( i = 0; i < h; ++i ) foo[i] = new int[3];  

And, by extension, for a 3D array:

  int i,j;int ***foo = new (int**)[w];   // create the vector of columnsfor( i = 0; i < w; ++i ){   foo[w] = new (int*)[h];     // create a column   for( j = 0; j < h; ++j )      foo[w][h] = new int[3];  // create a cell}  

Now, foo points to an array of columns, foo[x] points to column x and foo[x][y] points to element (x,y).

Of course you have to be careful with your cleanup or you will get a nifty memory leak.

  int i,j;for ( i = 0; i < w; ++i ){  for( j = 0; j < h; ++j )    delete[] foo[i][j];     // delete a cell  delete[] foo[i];          // once done, delete the column}delete[] foo;               // delete the vector of columns  

Yet another solution, is to use std::vector, there is much less bookkeeping (and after all, that is what they have been designed for).

  #include <vector>typedef int cell_t[3];typedef std::vector<cell_t> column_t;typedef std::vector<column_t> array_t;int i;array_t foo(w);         // make an array of w columnsfor( i = 0; i < w; ++i )     foo[i].resize(h);   // set the size of each column to be h.  

calling std::vector::resize() to make the array bigger or smaller will take care of the memory problems.



Edited by - Fruny on November 16, 2001 4:47:00 AM