constant elements vs. constant array

Question

Everybody knows how to declare an array with constant elements:

const int a[10];

Apparently, it is also possible to declare an array that is itself constant, via a typedef:

typedef int X[10];
const X b;

From a technical and a practical standpoint, do a and b have the same type or different types?

An array can have constant elements, the size of array is compile time constant.. but what does a constant array even mean? M curious. — SandBag_1996, May 26 '16 at 14:54
A similar question: http://stackoverflow.com/questions/31412972/what-does-int-const-a5-really-mean — Blagovest Buyukliev, May 26 '16 at 15:01
@BlagovestBuyukliev It's not a similar question, I think it is a duplicate. But the answerer seems to imply that the array itself is "const", which is not true. — uh oh somebody needs a pupper, May 26 '16 at 15:26

score 4 · Answer 1 · answered May 26 '16 at 15:01

They have the same type, though clang will print them differently. Since the array itself cannot be const, it is said to "fall through".

For the non-typedefed version, clang prints the type as const int [10]. For the typedef version, it prints int const[10]. Both of those are equivalent.

Coliru in action.

score 4 · Accepted Answer · answered May 26 '16 at 15:20

Apparently, it is also possible to declare an array that is itself constant

Nope. In N1256, §6.7.3/8:

If the specification of an array type includes any type qualifiers, the element type is so-qualified, not the array type.¹¹⁸⁾

Then footnote 118 says:

Both of these can occur through the use of typedefs.

score 1 · Answer 3 · answered May 26 '16 at 15:11

Ran some tests:

typedef int X[10];

#define typeinfo(t) _Generic((t), \
  X: "int X[10]", \
  int: "int", \
  int *: "int *", \
  const int *: "const int *", \
  default: "not" \
  )

int main(void) {
  const int a[10];
  const X b;

  printf("%zu\n", sizeof (const int *));
  puts(typeinfo(a));
  puts(typeinfo(b));
  printf("%zu\n", sizeof a);
  printf("%zu\n", sizeof b);
  puts(typeinfo(a[0]));
  puts(typeinfo(b[0]));
  printf("%zu\n", sizeof a[0]);
  printf("%zu\n", sizeof b[0]);
}

Output

4
const int *
const int *
40
40
int
int
4
4

Both have the same size and both convert to the same type when passed as an argument. Neither is a pointer. Both are arrays.

By code analysis and testing - they are the same type.

They are also the same type by definition! – Ian Abbott May 26 '16 at 15:36 — Ian Abbott, May 26 '16 at 15:36

constant elements vs. constant array

3 Answers3