uh............. how do we use ''new'' to make a multidimensional array? i thought it was something like this..

    

class stuff
{
  public:
    stuff(int x, int y)
    {      
      grid = new short int[3][x][y];    
    }
  protected:
    short int* grid;
};

int main(void)
{
  stuff foo(3, 5);
  return 0;
}   

    

plz help mattd The world is your message board - (edited to be politically correct)

Offhand, I dont recall. Nonetheless try this:

xs = (x size of array)
ys = (y size of array)

int (*grid)[3][];

grid = new int [3][xs*ys];

and then reference it as such:

grid[0][y*xs+x]



regards,

GeniX

How silly of me - im sure you figured it out, but:

you could also

grid[1][y*xs+x]

or

grid[2][y*xs+x]

..

the reason you probably cant do grid[0][x][y] is because the compiler wont know what the x size of the grid is if you use a new to allocate it.

anyone else who can shed more light is more than welcome as my knowledge of such things gets iffy when u have pointers to multi-d arrays (esp. pointers to multi-d arrays of pointers




regards,

GeniX

You can''t do it that way in C++. You can in Java, I believe.

To do it in C++, you have to do something like this:

    class stuff{    public:        stuff (int x, int y)        {            grid = new short**[3];            for (int i = 0; i < 3; i++)            {                  grid<i> = new short*[x];                for (int j = 0; j < x; j++)                {  grid[i][j] = new short[y];  }            }        }    private:        short ***grid;};    

Don''t forget to include a destructor (you''ll probably want to store x and y in stuff as well).

I think that''s how you do it... I''ve only ever done it for 2D arrays, though.

Hey, how about this - it compiles under VC++6.0

When creating the pointer to the array the computer can only *not* know the first dimension (ie: it doesnt have to know how many repetitions of the remaining dimensions there are to access it).

So we create a pointer to such an array, initialise it (with some hax0r conversion) and can then access it as such:

int (*grid)[][3];

grid = (int(*)[][3]) new int[5][3];

(*grid)[0][0] = 00;
(*grid)[1][0] = 01;
(*grid)[5][5] = 55;

delete( grid );

Makes sense to me. Compiles and runs with no errors. So im *assuming* there is no problem with it.




regards,

GeniX

MAN where are my brains today... not with me im sure.

in my code you will spot a reference to 5,5 ... this doesnt exist in the array i created. Who knows why it didnt GPF under Win2k....



regards,

GeniX

    int xs;int ys;xs = whateverYouWant;  // can be entered by userys = whateverYouWant2; // can be entered by userint** array;array = new int[xs];for (int i = 0; i < xs; i++){   *array<i> = new int[ys];}    

now just access it like you would a normal array


"Now go away or I shall taunt you a second time"
- Monty Python and the Holy Grail
themGames Productions

thnks geniX - and you''re from NZ too! nice........

oh and you ncsu121978


The world is your message board - (edited to be politically correct)