passing 2D array of ptrs as parameter

This is a discussion on passing 2D array of ptrs as parameter within the C++ Programming forums, part of the General Programming Boards category; OK, say I have this Code: #include <iostream> #include <string> using namespace std; struct MyType { double x; int y; ...

  1. #1
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    passing 2D array of ptrs as parameter

    OK, say I have this
    Code:
    #include <iostream>
    #include <string>
    using namespace std;
    
    struct MyType
    {
    	double x;
    	int y;
    };
    
    int main()
    {
    	MyType t1;
    	MyType t2;
    	MyType t3;
    	MyType t4;
    	MyType* pMT = &t1;
    	MyType* pMT2 = &t2;
    	MyType* pMT3 = &t3;
    	MyType* pMT4 = &t4;
    	MyType* Map[2][2] = {{pMT, pMT2},
    						 {pMT3, pMT4}};
    	return 0;
    }
    How would the prototype for a function, that would take the Map array as an argument, look like? Like for example if I just wanted a function that printed all the values in the array
    Code:
    void PrintArray(??);
    And one quick thing, would it be possible to have a pointer that points to the Map array that I can use to access everything? Or am I going in the totally wrong direction with this whole thing here?
    Thanks in advance,
    -Psycho

  2. #2
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    Thanks for the help Salem

    Would there be a way to do the function parameter say if the program didn't know the dimensions of the multi-dimensional array? So that you could maybe pass:
    MyType* Map[2][2]
    or
    MyType* World[3][5]

    this is where I'm stuck, I can't think of any way to do that in my knowledge, because I know when you define a multi-dimensional array you can leave the first [] empty but you have to define the second set.

    Well anywho, thanks in advance,
    -Psycho

  3. #3
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    I'm curious why Salem declared the function prototype with both dimensions, since the leftmost dimension is not part of the type of an array. The type of your array, as you apparently realize, is:

    MyType*[][2]

    But then you seem to forget about the type when you ask if you can pass a 3 x 5 array to the function. What's the type of a 3x5 array? Is it the same type as your function parameter?

    "Would there be a way to do the function parameter say if the program didn't know the dimensions of the multi-dimensional array?"

    Your program can't store data in an array unless it knows the dimensions beforehand. At some point, the program has to set aside the memory, and you have to tell it how much to set aside. You can wait until run time to dynamically declare a two dimensional array depending on some input from the user, but before you can send the array to the function, you have to create it, and at that point your program knows the size of the array. That may be what you are after. In that case, your function parameter would be of type:

    MyType**

    which may look strange, but it just means it's a pointer to a location in memory where there is an array of pointers. To dynamically create a two dimensional array, you need to do this:
    Code:
    MyType** Map= new MyType*[first_dimension];
    for(int i=0; i<first_dimension; i++)
    {
        my_array[i] = new MyType[second_dimension];
    }
    
    ...
    ...
    PrintArray(Map);
    To release the memory you need to do this:
    Code:
    for(int p=0; p<first_dimension; p++)
    {
         delete [] Map[p];
    }
    delete [] Map;
    Last edited by 7stud; 06-13-2003 at 01:03 PM.

  4. #4
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    Allright thank you both

    As for newbie, I'm only one in the area of pointers. I know how to pass arrays to functions, just not how to do it when the width is undetermined.

    Well I'll try these methods out, and I'll stop bugging you with it
    Thanks again,
    -Psycho

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