pointers to 2d arrays

I have a class that works something like this:

Code:

---

class bleh {
public:
      char *charArray;
      bleh(int size);
private:
};

bleh::bleh(int size) {
    charArray = new char[size];
}

---

then I realized that I wanted charArray to be 2d
however,

Code:

---

class bleh {
public:
      char *charArray;
      bleh(int size1, size2);
private:
};

bleh::bleh(int size1, int size2) {
    charArray = new char[size1][size2];
}

---

doesn't werk

the only thing that I can get to work is somethint like:

Code:

---

class bleh {
public:
      char charArray[100][100];
      bleh(int size);
private:
};

bleh::bleh(int size) {
}

---

Keeps yelling about either (based on small changes on my part):
A) doesn't like pointer
B) can't cast
C) doesn't like that a 2d array have variables for it's size

Any clues how it can be done?

Mabye this might help some.

Code:

---

class bleh {
private:  // might want it private
      char charArray[100][100];
      bleh(int size);
public: //Might want this public
};

bleh::bleh(int size) {
}

---

This might not help much but i changes private and public. Not sure if i can help you with the coding much. Ill try.

Or you can try something like this:

Code:

---

  char** array;
  const unsigned ROWS = 10;
  const unsigned COLS = 10;

  // allocate memory
  array = new char*[ROWS];

  for (i = 0; i < ROWS; ++i)
      array[i] = new char[COLS];
...

  // clean up memory
  for (i = 0; i < ROWS; ++i)
      delete array[i];

  delete array;

---

ah people are up this late ^_^ good

Salem's answer is what I'm wanting, thnx

I did a search and didn't find that post =/
srry

You can use one dimentional array then deal with it as a two dimentional array.

int *array;

array = new int[columns*rows]

so it's like putting the rows next to each other.

You can access each element by the index array[x, y];
like this;

array[x, y]; is array[ columns * x + y ];
where columns is the number of columns in the array.

You can put it in a function it will make it easier for you.
The function will return a pointer.

Do you get it.