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.

+1  A: 

Yes. Take a look at this page.

Jeremy Privett

Simply put:

An array is a pointer to a vector in memory. So, an array of arrays is a pointer to a pointer to a vector in memory.


In a near Galaxy you have
char a[100][100];
giveMeTheAnswer( a );

The event horizon must accept that multiverse
void giveMeTheAnswer( char** pMatrix ) {
pMatrix[0][0] = '4';
pMatrix[0][1] = '2';
yeah that is totally wrong :) try pMatrix[1][0] = '4'; :)
Johannes Schaub - litb
Wow that is so wrong. You completely confused arrays of arrays and arrays of pointers. They are completely different.
The type of pMatrix is `char (*)[100]` (pointer to 100-element array of char), not `char **`. Secondly, *arrays are not pointers*. Arrays and pointers are *different* objects that, in some contexts, can be used interchangeably.
John Bode
+6  A: 

You can do this with any data type. Simply make it a double 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(sizeof(char) * 50); //50 chars

array[y][x].myint = 6;
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.
+2  A: 

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+j*Z+k] = ...;
func_3d(&arr[0][0][0], 10, 20, 30);
John Bode