original title: "Passing multidimensional arrays as function arguments in C"


In C can I pass a multidimensional array to a function as a single argument when I don't know what the dimensions of the array are going to be ?

In addition my multidimensional array may contain types other than strings.



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    You can do this with any data type. Simply make it a pointer-to-pointer:

    typedef struct {
      int myint;
      char* mystring;
    } data;
    data** array;

    But don't forget you still have to malloc the variable, and it does get a bit complex:

    int x,y,w,h;
    w = 10; //width of array
    h = 20; //height of array
    //malloc the 'y' dimension
    array = malloc(sizeof(data*) * h);
    //iterate over 'y' dimension
      //malloc the 'x' dimension
      array[y] = malloc(sizeof(data) * w);
      //iterate over the 'x' dimension
        //malloc the string in the data structure
        array[y][x].mystring = malloc(50); //50 chars
        array[y][x].myint = 6;
        strcpy(array[y][x].mystring, "w00t");

    The code to deallocate the structure looks similar - don't forget to call free() on everything you malloced! (Also, in robust applications you should check the return of malloc().)

    Now let's say you want to pass this to a function. You can still use the double pointer, because you probably want to do manipulations on the data structure, not the pointer to pointers of data structures:

    int whatsMyInt(data** arrayPtr, int x, int y){
      return arrayPtr[y][x].myint;

    Call this function with:

    printf("My int is %d.\n", whatsMyInt(array, 2, 4));


    My int is 6.

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    Pass an explicit pointer to the first element with the array dimensions as separate parameters. For example, to handle arbitrarily sized 2-d arrays of int:

    void func_2d(int *p, size_t M, size_t N)
      size_t i, j;
      p[i*N+j] = ...;

    which would be called as

    int arr1[10][20];
    int arr2[5][80];
    func_2d(&arr1[0][0], 10, 20);
    func_2d(&arr2[0][0], 5, 80);

    Same principle applies for higher-dimension arrays:

    func_3d(int *p, size_t X, size_t Y, size_t Z)
      size_t i, j, k;
      p[i*Y*Z+j*Z+k] = ...;
    func_3d(&arr[0][0][0], 10, 20, 30);

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    You can declare your function as:

    f(int size, int data[][size]) {...}

    The compiler will then do all pointer arithmetic for you.

    Note that the dimensions sizes must appear before the array itself.

    GNU C allows for argument declaration forwarding (in case you really need to pass dimensions after the array):

    f(int size; int data[][size], int size) {...}

    The first dimension, although you can pass as argument too, is useless for the C compiler (even for sizeof operator, when applied over array passed as argument will always treat is as a pointer to first element).

  • Translate
    int matmax(int **p, int dim) // p- matrix , dim- dimension of the matrix 
        return p[0][0];  
    int main()
       int *u[5]; // will be a 5x5 matrix
       for(int i = 0; i < 5; i++)
           u[i] = new int[5];
       u[0][0] = 1; // initialize u[0][0] - not mandatory
       // put data in u[][]
       printf("%d", matmax(u, 0)); //call to function
       getche(); // just to see the result