Question

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?

Besides, my multidimensional array may contain types other than strings.

Answer 1

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:

//initialize 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 for(y=0;y<h;y++){   //malloc the 'x' dimension   array[y] = malloc(sizeof(data) * w);    //iterate over the 'x' dimension   for(x=0;x<w;x++){     //malloc the string in the data structure     array[y][x].mystring = malloc(50); //50 chars      //initialize     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)); 

Output:

My int is 6.