not passing rows count of 2d arrays to a function in c

Question

can i do some calculation using the sizeof operator in c to get the rows count inside a function who receives as a parameter only a two dimensional array without any information about rows or columns count in this passed array?

void main(void){
    int A[5][6];
    int columnsC = sizeof(A[0]) / sizeof(int);
    int rowsC = ( sizeof(A)/sizeof(int) ) / columnsC;
    printf("rows: %d\ncolumns: %d\n", rowsC, columnsC);

    getch();
}

the previous code worked for me, but i can't find a helpful resource to know how to apply this on a function of the following header:

void fun(int arr[][10])

i"m studying a course on Programming Languages Principles, and some times think of things differently and look for learning resources, and this helps a lot, but not this time apparently.

i admire any help.

Answer 1

No. There's no way find the size of an array when passed to a function. Array decays into a pointer when passed to a function.

This:

void fun(int arr[][10])

is actually syntatic sugar for:

void fun(int (*arr)[10])

Instead you can pass the size as a thrid argument or size as an element of the array itself.

Answer 2

I do not believe this is possible, unless you either pass a length, or you specifically allocate one extra row to the array. This extra row can contain a null character. In the function fun, then you can use a while loop to cycle through the array until you hit the null value.

int i = 0;
while (1) {
    if(arr[i]==NULL){
        //exits when hits end    
        break;
    }
    //do stuff with arr[i]
    i++;
}