Passing Two Dimension array to a function as a single pointer - c

Question

I need to pass Two Dimension array to a function as a single pointer. There are different types of approaches are there but due to some constraints(CodeGeneration), I want to pass a single pointer only. I have macros which contain the size of each dimension. I implemented the following way but I am not sure it will work fine for N dimensions also
#define size_1D 3
#define size_2D 3
void fun(int *arr)
{
int i,total_size = size_1D* size_2D;
for(i = 0; i < total_size ; i++)
{
int value = arr[i];
}
}
int main()
{
int arr[size_1D][size_2D] = {{1,2,7},{8,4,9}};
fun(&arr[0][0]);
}
Any loophole is there if I followed the above approach?

Answer 1

void fun(int (*arr)[3]);
or exactly equivalent, but maybe more readable:
void fun(int arr[][3]);
arr is a pointer to two dimensional array with 3 rows and 3 columns. arr decayed to a pointer has the type of a pointer to an array of 3 elements. You need to pass a pointer to an array of 3 elements. You can access the data normally, using arr[a][b].
#define size_1D 3
#define size_2D 3
void fun(int arr[][3])
{
for(int i = 0; i < size_1D ; i++) {
for(int j = 0; j < size_2D ; j++) {
int value = arr[i][j];
}
}
}
int main()
{
int arr[size_1D][size_2D] = {{1,2,7},{8,4,9}};
fun(arr);
}
You can specify the sizes as arguments and use a variable length array declaration inside function parameter list. The compiler will do some job for you.
#include <stdlib.h>
void fun(size_t xmax, size_t ymax, int arr[xmax][ymax]);
// is equivalent to
void fun(size_t xmax, size_t ymax, int arr[][ymax]);
// is equivalent to
void fun(size_t xmax, size_t ymax, int (*arr)[ymax]);
void fun(size_t xmax, size_t ymax, int arr[xmax][ymax])
{
for(int i = 0; i < xmax ; i++) {
for(int j = 0; j < ymax ; j++) {
int value = arr[i][j];
}
}
}
int main()
{
int arr[3][4] = {{1,2,7},{8,4,9}};
fun(3, 4, arr);
}
#edit
We know that the result of array subscript operator is exactly identical to pointer dereference operator of the sum:
a[b] <=> *(a + b)
From pointer arithmetic we know that:
type *pnt;
int a;
pnt + a = (typeof(pnt))(void*)((uintptr_t)(void*)pnt + a * sizeof(*pnt))
pnt + a = (int*)(void*)((uintptr_t)(void*)pnt + a * sizeof(type))
And that the array is equal to the value to the pointer to the first element of an array:
type pnt[A];
assert((uintptr_t)pnt == (uintptr_t)&pnt[0]);
assert((uintptr_t)pnt == (uintptr_t)&*(pnt + 0));
assert((uintptr_t)pnt == (uintptr_t)&*pnt);
So:
int arr[A][B];
then:
arr[x][y]
is equivalent to (ignore warnings, kind-of pseudocode):
*(*(arr + x) + y)
*( *(int[A][B])( (uintptr_t)arr + x * sizeof(int[B]) ) + y )
// ---- x * sizeof(int[B]) = x * B * sizeof(int)
*( *(int[A][B])( (uintptr_t)arr + x * B * sizeof(int) ) + y )
// ---- C11 6.5.2.1p3
*( (int[B])( (uintptr_t)arr + x * B * sizeof(int) ) + y )
*(int[B])( (uintptr_t)( (uintptr_t)arr + x * B * sizeof(int) ) + y * sizeof(int) )
// ---- *(int[B])( ... ) = (int)dereference( ... ) = *(int*)( ... )
// ---- loose braces - conversion from size_t to uintptr_t should be safe
*(int*)( (uintptr_t)arr + x * B * sizeof(int) + y * sizeof(int) )
*(int*)( (uintptr_t)arr + ( x * B + y ) * sizeof(int) )
*(int*)( (uintptr_t)( &*arr ) + ( x * B + y ) * sizeof(int) )
// ---- (uintptr_t)arr = (uintptr_t)&arr[0][0]
*(int*)( (uintptr_t)( &*(*(arr + 0) + 0) ) + ( x * B + y ) * sizeof(int) )
*(int*)( (uintptr_t)( &arr[0][0] ) + ( x * B + y ) * sizeof(int) )
*(int*)( (uintptr_t)&arr[0][0] + ( x * B + y ) * sizeof(int) )
// ---- decayed typeof(&arr[0][0]) = int*
*( &arr[0][0] + ( x * B + y ) )
(&arr[0][0])[x * B + y]
So:
arr[x][y] == (&arr[0][0])[x * B + y]
arr[x][y] == (&arr[0][0])[x * sizeof(*arr)/sizeof(**arr) + y]
On a sane architecture where sizeof(uintptr_t) == sizeof(size_t) == sizeof(int*) == sizeof(int**) and etc., and there is no difference in accessing data behind a int* pointer from accessing data behind int(*)[B] pointer etc. You should be safe with accessing one dimensional array when using a pointer to the first array member, as the operations should be equivalent ("safe" with exception for out-of-bound accesses, that's never safe)
Note, that this is correctly undefined behavior according to C standard and will not work on all architectures. Example: there could be an architecture, where data of the type int[A] are stored in different memory bank then int[A][B] data (by hardware, by design). So the type of the pointer tells the compiler which data bank to choose, so accessing the same data with the same to the value pointer, but with different pointer type, leads to UB, as the compiler chooses different data bank to access the data.