Is it possible to determine amount of rows of a 2-d array passed in a function

[hugo2x]

I am trying to make a function that accepts a an array by reference and determines the amount of rows in the 2-d array.
Is there a way to do this? (I do know the columns)

void printer(int [][3]);
int main(void)
{
	int array1[][3]={{ 5,9, 0},
			{14,7,13},
	                 { 2,10,8}};
	int array2[][3]={{ 5,9, 0},
	                 {14,7,13},
                          {55,10,9}
	                  { 2,10,8}};
	
	printer(array1);
        printer(array2);
}

void printer(int ary[][3])
{
	int rows = ////// some way to calculate 3 from first instance and 4 for second instance, I have tried sizeof(array)/sizeof(int) but it only calculates the size of the address of array(0Xff244da) */
	
}

Any help would greatly be appreciated!

[King Mir]

If by rows you mean the first dimension of the array, that is omitted from the deceleration, then there is no way given the prototype. You can do this:

void printer(int (*ary)[4][3])//pointer to 4x3 array
{
        int cols = sizeof(**ary)/sizeof(***ary);
        /*size of each row devided by the size of each element in each row. (sizeof(***ary)==sizeof(int))
        cols should equal 3.*/
	int rows = sizeof(*ary)/sizeof(**ary); 
        /*This computes the size of the whole array divided by the size of each row.
        rows should equal 4.*/
	(*ary)[1][1]=5;/*this is how to access the array. */
        ary[0][1][1]=5;/*this is also valid */
}

It's not as ugly in C++.

[Adak]

Could you just pass both dimensions? There had to be a variable for the rows when you put the data into the array.