invalid pointer conversion

This is a discussion on invalid pointer conversion within the C++ Programming forums, part of the General Programming Boards category; Hello, Consider following code: Code: void foo(int **arr); int main() { int array[10][10]; int **ptr; ptr = array; // error ...

  1. #1
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    invalid pointer conversion

    Hello,
    Consider following code:

    Code:
    void foo(int **arr);
     
    int main()
    {
        int array[10][10];
        int **ptr;
        
        ptr = array; // error here
        
        foo(array);  // here aswell
        printf("%d", array); 
     
        getchar();    
        return  0; 
    }
    
    
    void foo(int **arr) 
    {
         printf("Hello\n");
    }
    What's wrong with this? I've tried same code in C, it works except that complier throws warning about incompatible pointers, while in C++ it's an error - any ideas?

  2. #2
    Registered User Noir's Avatar
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    A 2D array isn't a pointer to a pointer, it's an array of pointers. These are the two types you're trying to mix, and they're not compatible:
    Code:
    int **
    int (*)[10]

  3. #3
    Registered User pronecracker's Avatar
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    And how would you declare a pointer to an array of arrays of four bytes? If you don't know how many groups of four bytes there will be?

  4. #4
    and the hat of wrongness Salem's Avatar
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    > void foo(int **arr);
    Would be
    void foo(int (*arr)[10]);
    or more simply
    void foo(int arr[ ][10]);
    or even just copy/paste for the lazy
    void foo(int array[10][10]);
    The definition of course follows the prototype.

    The pointer in main() follows the same logic, say
    int (*ptr)[10] = array;
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  5. #5
    Registered User Noir's Avatar
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    It depends on how the groups were defined to begin with. If it's a 2D array that you just don't know the size of the first dimension you can do the same thing:
    Code:
    char (*p)[4];
    It's probably not an array though if you don't know the size of one of the dimensions. It's probably something dynamically allocated, and this has the best chance of working:
    Code:
    char **p;
    because this is the most common way to allocate memory for a 2D array:
    Code:
    char **p = new char*[n];
    
    for ( int i = 0; i < n; i++ ) {
      p = new char[4];
    }
    But if the memory is allocated like this the first way is the one you want:
    Code:
    int (*p)[4] = new int[n][4];

  6. #6
    Cat without Hat CornedBee's Avatar
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    because this is the most common way to allocate memory for a 2D array:
    Not really - there better things to do with memory than wasting it on pointers and allocation overhead.

    The most common way is to allocate a 1D array and do index calculations.
    Last edited by CornedBee; 04-13-2007 at 10:23 AM. Reason: Rephrasing
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  7. #7
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    What if, say I got 2 local arrays:
    Code:
    int array1[10][10];
    int array2[100][100];
    ...and I want to create a function that could take any of them as an argument?
    Well, if I call malloc to allocate memory for those arrays, then it's no problem. But what if I want to keep it like that, and making foo() able to take either array1 or array2?

  8. #8
    Registered User whiteflags's Avatar
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    One way is to flatten the array and then pass the first element to your function.

    void foo ( int *array, size_t total_x, size_t total_y );

    ...

    foo( &array1[0][0], total_x, total_y );

    You have to do some arithmatic like this_x * total_y + index to access it correctly, but it shouldn't be too much of a problem.
    Last edited by whiteflags; 04-13-2007 at 12:25 PM.

  9. #9
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    Quote Originally Posted by jimzy View Post
    What if, say I got 2 local arrays:
    Code:
    int array1[10][10];
    int array2[100][100];
    ...and I want to create a function that could take any of them as an argument?
    Well, if I call malloc to allocate memory for those arrays, then it's no problem. But what if I want to keep it like that, and making foo() able to take either array1 or array2?
    This comes up a lot in interfacing with apis. For example, you have to do it to use libpng since it deals with arbitrary sized 2d arrays. You have to do some preprocessing.

    Code:
    {
       int array1[10][10];
       int * array1_rows[10];
       
       for (i = 0; i < 10; ++i) {
          array1_rows[i] = &array1[i][0];
       }
       
       // Now, array1[x][y] == array1_rows[x][y], but array1_rows can be passed to int** functions as a 2d array.
    }
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