What is the advantage of a pointer to the base address of an array instead of pointer to the first element?

Question

Following is a code snippet showing a pointer to the base of an array and a pointer to the element;

1st case:

/* Pointer to an array of 10 integers */
int arr[10]
int (*ptr)[10];
ptr = & arr; //points to the base address of an array of 10 integers
for (i = 0; i < 10; i++) //prints the values
    printf("%d\n", *(*ptr + i)); 

2nd case:

/* Pointer to an array element*/
int arr[10];
int *ptr;
ptr = arr;  //Points to the first element of an array
for (i = 0; i < 10; i++) //prints the values
{
    printf("%d\n", *ptr);
    ptr ++;
}

I understood the difference in accessing the values in both cases. However I didnot understand what is the advantage of using the first case i.e. pointer to an array of 10 integers even if the second case looks simple.

Answer 1

In your particular example

int arr[10]
int (*ptr)[10];
ptr = & arr; //points to the base address of an array of 10 integers
for (i = 0; i < 10; i++) //prints the values
    printf("%d\n", *(*ptr + i));  //NOTE: same as `ptr[0][i]`

there's no advantage or at least no advantage being utilized.

*(*ptr+i) with int (*ptr)[10]; will/should generate the same assembly as *(ptr+i) would with int ptr[10]; (Note: some/many might find ptr[0][i] and ptr[i] respectively as more readable renderings of these expressions)

Example:

int get_nth(int (*X)[10], int N) { return *(*X+N); }
int get_nth_(int *X, int N) { return *(X+N); }

x86_64 output (gcc -O3 or clang -O3):

get_nth:                                # @get_nth
        movsxd  rax, esi
        lea     rax, [rdi + 4*rax]
        ret
get_nth_:                               # @get_nth_
        movsxd  rax, esi
        lea     rax, [rdi + 4*rax]
        ret

https://gcc.godbolt.org/z/up7HXc

If int (*ptr)[10] were derived from a multidimensional array as in

int multidim[5][10]; //array 10 or array 5 of int
int (*ptr)[10]=&multidim[1];

you could use the first index to jump the pointer in increments of 10*sizeof(int) in addition to using the second one to jump in increments of sizeof(int) (as with a plain int*).

In a standalone example (i.e., where the 10-int block isn't part of a multidimensional array), probably the only "advantage" of int(*)[10] is that it retains the sizeof information (even across a function call boundary), and you could conceivably use this for explicit bounds-checking.

Bounds-checking example:

#include <stdio.h>

#define CHECKED_SUBSCRIPT(PtrArray, Index) (*(PtrArray))[BOUNDCHECK(Index,ARRCNT(*(PtrArray)))] /*{{{*/
    #include <assert.h>
    static inline size_t BOUNDCHECK(size_t Index, size_t Bound)
    {
        assert(Index <  Bound);
        return Index;
    }
        //sizeof(A) or void (=>error) if A isn't an array
        #define ARRSZ(A) (+ sizeof(A) + 0* /*{{{*/\
                     _Generic((__typeof(&(A)[0])*)0, __typeof(__typeof(A)*):(void)0,default:0) ) /*}}}*/
    #define ARRCNT(A) (ARRSZ(A)/sizeof((A)[0])) /*}}}*/
int main()
{
    int arr[10]={0,2,4,6,8,10,12,14,16,18};
    int (*ptr)[10] = &arr;
    for (int i = 0; i < 20; i++){
        printf("%d\n", CHECKED_SUBSCRIPT(ptr,i));
    }
}

Output:

0
2
4
6
8
10
12
14
16
18
a.out: boundschecking.c:7: BOUNDCHECK: Assertion `Index < Bound' failed.

Answer 2

There is no advantage. The types of the pointers to access the elements of array arr are just different.

In the first case, ptr is of type int (*)[10] - pointer to array of 10 int.

In the second case, ptr is of type int * - pointer to int.

---

ptr = arr; - arr decays to a pointer to the first element of arr (type int*).

ptr = &arr; - &arr is a pointer to an array of 10 elements to int (type int (*)[10]).

---

Side note:

If you would increment ptr in the first example, the pointer will point to a memory location past the array arr. Dereferencing that pointer would invoke undefined behavior at least in C:

When an expression that has integer type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the pointer operand points to an element of an array object, and the array is large enough, the result points to an element offset from the original element such that the difference of the subscripts of the resulting and original array elements equals the integer expression. In other words, if the expression P points to the i-th element of an array object, the expressions (P) + N (equivalently, N + (P)) and (P) - N (where N has the value n) point to, respectively, the i + n-th and i − n-th elements of the array object, provided they exist.

Moreover, if the expression P points to the last element of an array object, the expression (P) + 1 points one past the last element of the array object, and if the expression Q points one past the last element of an array object, the expression (Q) - 1 points to the last element of the array object. If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined. If the result points one past the last element of the array object, it shall not be used as the operand of a unary * operator that is evaluated.

Source: C18, 6.5.6/8

For C++: Dereferencing one past the end pointer to array type

While incrementing ptr in the second example just let ptr point to the following char element of array arr.