Thread: Multi-Demensional Arrays on Heap

  1. #31
    &TH of undefined behavior Fordy's Avatar
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    Originally posted by The Dog
    Doh, my bad!

    I didn't see this : new Room[20];

    I assumed it was : new Room;

    I'm sorry Enmeduranki, but i was following my code instead of Fordy's.


    ::Throws the dog a bone::

  2. #32
    Carnivore ('-'v) Hunter2's Avatar
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    wow, this is a pretty heated argument... I agree with Enmeduranki. 20 arrays of 50 chars each makes 1000 chars in total, not 50. This is because of the ancient and nearly-forgotten mathematics technique calles "multiplication", which states that "20 X 50 = 1000". And an array of 20 arrays of 20 rooms each is 20 arrays of 20 rooms each, or 400 rooms in total.

    In the previous code we had 20 pointers, so we created 20 Room objects on the heap.
    So thats 20 pointers to arrays of 20. And each pointer is turned into an array of 20 arrays of 20 rooms each.

    As for the concept of "20 pointers = 20 rooms", lets look at this: If you have 1 array of 20 rooms, then you have 1 pointer, and 20 rooms. So "1 pointer = 20 rooms". so "20 pointers = 20 X 20 rooms".
    Just Google It. √

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  3. #33
    Registered User The Dog's Avatar
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    btw, i still stand by my code!

    >>
    //dynamically allocated an array of pointers
    //this isn't necessary if one of your dimensions
    //is a compile time constant
    int** array = new int*[20];

    //allocate an array of 20
    //integers (yes, 20 and they're integers you know)
    //to each of the previously allocated pointers

    array[0]=new int[20];
    array[1]=new int[20];
    array[2]=new int[20];
    array[3]=new int[20];
    array[4]=new int[20];
    array[5]=new int[20];
    array[6]=new int[20];
    array[7]=new int[20];
    array[8]=new int[20];
    array[9]=new int[20];
    array[10]=new int[20];
    array[11]=new int[20];
    array[12]=new int[20];
    array[13]=new int[20];
    array[14]=new int[20];
    array[15]=new int[20];
    array[16]=new int[20];
    array[17]=new int[20];
    array[18]=new int[20];
    array[19]=new int[20];

    There's quite a few 20s there.
    <<
    Yet array only holds 20 arrays, each which hold 20 chars!

  4. #34
    Registered User GrNxxDaY's Avatar
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    i think i made it work also, but with one less * (asterik).
    See anything wrong people??..
    Code:
    ROOM** createRooms()
    {    
      // for a 20x5 array of ROOMS
    
      ROOM** room;
      room = new ROOM*[5];    //room now holds the address to the first of 5 pointers to ROOM objects
    
      for (int i=0; i<5; i++)
        room[i] = new ROOM[20];    //room[0] holds address of first of 20 ROOMS
    
      return room;    //return a ROOM**
    }
    
    //..somewhere in main()
    
    for (int i = 5; i<5; i++)
      delete [] room[i];
    room = 0;
    Last edited by GrNxxDaY; 08-13-2002 at 01:53 PM.
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  5. #35
    Carnivore ('-'v) Hunter2's Avatar
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    *forget the post above*
    Ah, the traditional old "need glasses" problem...
    Just Google It. √

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  6. #36
    and the hat of int overfl Salem's Avatar
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    Man, what a lot of repies

    > ROOM** createRooms()
    This is fine - and if you understand it, use it. There is just one small point to add

    for (int i = 0; i<5; i++)
    &nbsp;&nbsp;delete [] room[i];
    delete [] room; // add this
    room = 0;

    Like I said, if you're happy you understand it, go with it, and ignore my more unusual pointers




    The rest of this is an explanation of my previous posts. Pointers to arrays are a bit unusual, and take some getting used to.

    > 1)what does the Lvalue on the assignment statment mean? the [20] slightly confuses me.
    Ok, there are two very similar looking declarations which mean two very different things.

    ROOM (*room)[20];
    This is just ONE pointer, but it is a pointer to an array of 20 ROOMS
    See my previous post on how to initialise this

    ROOM *room[20];
    This is an array of 20 pointers, each of which points to an array of ROOMS. This is initialised using
    for(x = 0;x < 20;x++) room[x] = new ROOM[20];

    Both give you the same room[20][20] structure by slightly different routes.

    > 2)how can i return the room pointer back to main? what return type do i use?
    Now this is where it gets a bit tricky.
    Code:
    #include <iostream>
    using namespace std;
    
    class ROOM {
        int    id;
    };
    
    // This is the hard way
    // pointers to arrays have a really weird looking notation
    // which gets even weirder when you try and return such a pointer
    // from a function.
    // trust me, you really dont want to do it this way
    ROOM (* hard_way( void ) )[20] {
        ROOM (*p)[20];
        p = new ROOM[20][20];
        return p;
    }
    
    // This is much better, we use a typedef to create a new
    // name for the row of the array
    // ROOMROW is an array of 20 ROOMS
    typedef ROOM ROOMROW[20];
    
    // See how much better this looks, even though
    // it is identical to the previous example
    ROOMROW *easy_way ( void ) {
        ROOMROW   *p;
        p = new ROOMROW[20];	// 20 lots of ROOM[20]
        return p;
    }
    
    int main() {
        ROOM     (*p)[20] = hard_way( );
        ROOMROW   *q = easy_way( );
    
        // we now have p[10][20] and q[10][20]
        for ( int r = 0 ; r < 20 ; r++ ) {
            for ( int c = 0 ; c < 20 ; c++ ) {
                p[r][c] = q[r][c];
            }
        }
    
        delete[] p;
        delete[] q;
        return 0;
    }

  7. #37
    Registered User GrNxxDaY's Avatar
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    Okay all - thanks so much! I'm pretty sure that it works now. I guess it was so hard for me cause I have trouble looking at code and understanding what it does.
    Well if any of you are interested and want some gratification (is that the right word?) of your good services, here's the code/program that i've been working on - you all can download it if you want just to see it.. err.. well.. umm.. whatever.

    Thanks again!!
    (will post the program once i restart computer)
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  8. #38
    Registered User GrNxxDaY's Avatar
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    here it is so far
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  9. #39
    Registered User GrNxxDaY's Avatar
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    Passing Constant Pointers to Pointers

    I have a pointer to a pointer and I'd like to pass that to a function, and have that function accept is as a constant thingy, so that function can only call const member functions (does that make sense?) I can't seem to get it to work though.. here's a little sniplett..
    Code:
    int main()
    {
      //...
    
      ROOM** roomPtr; 
      roomPtr = createRooms();   // works great
    
      functionOne(roomPtr);
     
     //...
    }
    
    
    functionOne(const ROOM** roomPtr)
    {
      //now this roomPtr should only call const member functions of ROOM type
    }
    BTW, i get get it to work easily enough for a regular pointer (like functionTwo(const MYCLASS* classPtr)...
    AOL: GrNxxDaY
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  10. #40
    and the hat of int overfl Salem's Avatar
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    I think you need to apply the const to both levels of indirection

    Like so
    Code:
    void functionOne(const ROOM* const* roomPtr)
    {
      //now this roomPtr should only call const member functions of ROOM type
    }
    This prevents you from modifying a roomPtr[i] with say
    roomPtr[i] = new ROOM

    And prevents you from modifying a data element with say
    roomPtr[i][j].fred = 1;

  11. #41
    Registered User GrNxxDaY's Avatar
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    thakns so much salem.. i'll try that out
    AOL: GrNxxDaY
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    If you think I may need help, please IM me.

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