I have a pointer which is meant to point to a two dimensional array like this: pointer[1][2] I know that the second dimension always is nine, but I don''t know the first dimension: pointer[?][9] I need to allocate memory dynamicaly for this. How do I do it? When I want to do this: int ArraySize = 10; int *pointer; pointer = new int[ArraySize][9]; The compiler tells me: error C2440: ''='' : cannot convert from ''short (*)[9]'' to ''short *'' Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast Any ideas how to do this? "Mr Sandman bring me a dream"

How about:

int ArraySize = 10;
int **pointer;
pointer = new int *[ArraySize];

for (int index=0; index < ArraySize; index++)
{
pointer[index] = new int [9];
}

Note: I'm just a beginner programmer; and I'm using this sort of call in a class (i.e. "this->pointer = new int *[ArraySize]") - so I don't know if you might need some pointer-dereferences ("*") in front of "pointer" in the above code... see what the compiler says.

Good luck!

--HB

P.S. In the above example, that's for a 10 by 9 array. ArraySize becomes the variable holding your first-dimension size.


Edited by - HBringer on November 9, 2000 6:17:38 PM

yeah that's how you do it
just though I'd add the delete

    int rows=6, cols=9;int** Array2D;Array2D = new int*[rows];for(int j=0;j<cols;j++)	Array2D[j] = new int[cols];//...for(int j=0;j<cols;j++)	delete[] Array2D[j];delete[] Array2D;  

i's don't show up in [] because it think you mean italics... guess I'll just use j's for now on

Edited by - Magmai Kai Holmlor on November 9, 2000 6:37:31 PM

Ooohhh, good catch! I almost forgot about those; but since all my stuff is OOP, I have those tucked away in Destructors and cleanup functions (outta sight, outta mind)... :-P

--HB

Uhhh, thanx guys.
Works perfectly. I was pretty busy in the last few days and so I couldn''t answer earlier.
Anyway: you were a great help once again.
Thanks!

"Mr Sandman bring me a dream"

wouldn''t it work to just do this:

  int iArraySize;int * lpIntArray;lpIntArray = new int[9 * iArraySize];  

Of course, you would have to do something like this to access the array 2-dimensionally: int iSomeInt = lpIntArray[X + Y * 9];