passing arrays/pointers to functions

let's say i have this code:

Code:

---

int main() {
int size = 6;
char ez[size][size];
//fill up 2d array ez with some stuff
  deal_with(ez,size);
}

void deal_with( ????? , int size) {



}

---

what kind of thing do i pass to the function? i've tried passing it as a 2d array, but since size is not predetermined, it can't guess the size of the array. i've tried a char** but it gives me an error. any ideas?

>i've tried a char** but it gives me an error.
Understandable. When someone sees that an array passed to a function can be passed as either

char array[]
or
char *array

They feel that a similar method can be taken with multidimensional arrays:

char array[][10]
or
char **array

But, in reality, two dimensional arrays are degraded into a pointer to an array:

char array[][10]
becomes
char (*array)[10]

In your case, the use of a structure would minimize headaches. Create the array in a structure, along with the size of each dimension and then allocate the required space at run-time. When you want to pass the array, all you have to do is pass the struct instance and not have to worry about the annoying syntax of 2D arrays as function parameters.

-Prelude

Not quite perfect, but it seems to work

Code:

---

#include <iostream.h>

void foo ( int size, void *p ) {
    int (*arr)[size] = (int(*)[size])p;
    cout << "Size is " << size << endl;
    int r, c;
    for ( r = 0 ; r < size ; r++ ) {
        for ( c = 0 ; c < size ; c++ ) {
            cout << arr[r][c] << " ";
        }
        cout << endl;
    }
}

int main ( ) {
    int r,c;

    int size = 5;
    int arr[size][size];
    for ( r = 0 ; r < size ; r++ ) {
        for ( c = 0 ; c < size ; c++ ) {
            arr[r][c] = r * c;
        }
    }
    foo ( size, arr );

    int size2 = 8;
    int arr2[size2][size2];
    for ( r = 0 ; r < size2 ; r++ ) {
        for ( c = 0 ; c < size2 ; c++ ) {
            arr2[r][c] = r * c;
        }
    }             
    foo ( size2, arr2 );

    return 0;
}

---

The obvious
void foo ( int size, int (*arr)[size] );
results in complaints about scope.

From a type point of view, each size would be a different function. You can't pass a [2][2] array to a function which is expecting a [3][3] array.

The downside is, the array is no longer checked, since its being passed to a void* parameter.

Hrrm I still have to look up some reference on structures (going to use this site's tutorial)
but - Prelude you're suggesting something like..

struct ...
{
int min=MIN;
int max=MAX;
char array[MIN][MAX];
}

yes?

Ohh never mind... I just went through the tutorial and finished looking through the message board more...

struct name{
char [5][25];
}

The thing is, you want the function to accept either a char** or a char*[]. You're just thinking too much :D

EZ:

Code:

---



void deal_with(char **strings, int max)
{ 
  int i = 0;
 
  printf("Printing From Within...");

  for(i = 0; i < max; i++)  printf(strings[i], i + 1); 
}






int main(int argc, char *argv[])
{
  int size = 100;

  char ez[ size ][ size ], *pointers[ size ];

  int i = 0;

  for(i = 0; i < size; i++) pointers[i] = strcpy(ez[i],"String #%i\n");

  deal_with( pointers, size );

  getch();
 
return 0;
}

---