C pointer : array variable

Question

I read this in my book (and many sources on the internet):

The array variable points to the first element in the array.

Answer 1

In most circumstances, an expression of array type ("N-element array of T") will be replaced with / converted to / "decay" to an expression of pointer type ("pointer to T"), and the value of the expression will be the address of the first element in the array.

So, assuming the declaration

int a[10];

the type of the expression a is "10-element array of int", or int [10]. However, in most contexts, the type of the expression will be converted to "pointer to int", or int *, and the value of the expression will be equivalent to &a[0].

The exceptions to this rule are when the array expression is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array in a declaration.

So, based on our declaration above, all of the following are true:

  Expression   Type         Decays to     Value
  ----------   ----         ---------     -----
           a   int [10]     int *         address of the first element of a
          &a   int (*)[10]  n/a           address of the array, which is the
                                            same as the address of the first
                                            element
       &a[0]   int *        n/a           address of the first element of a
          *a   int          n/a           value of a[0]
    sizeof a   size_t       n/a           number of bytes in the array 
                                            (10 * sizeof (int))
   sizeof &a   size_t       n/a           number of bytes in a pointer to 
                                           an array of int
   sizeof *a   size_t       n/a           number of bytes in an int
sizeof &a[0]   size_t       n/a           number of bytes in a pointer to int

Note that the expressions a, &a, and &a[0] all have the same value (address of the first element of a), but the types are different. Types matter. Assume the following:

int a[10];
int *p = a;
int (*pa)[10] = &a;

Both p and pa point to the first element of a, which we'll assume is at address 0x8000. After executing the lines

p++;
pa++;

however, p points to the next integer (0x8004, assuming 4-byte ints), while pa points to the next 10-element array of integers; that is, the first integer after the last element of a (0x8028).