Unsized array declaration in a struct

Question

Why does C permit this:

typedef struct s
{
  int arr[];
} s;

where the array arr has no size specified?

Answer 1

This is C99 feature called flexible arrays, the main feature is to allow the use variable length array like features inside a struct and R.. in this answer to another question on flexible array members provides a list of benefits to using flexible arrays over pointers. The draft C99 standard in section 6.7.2.1 Structure and union specifiers paragraph 16 says:

As a special case, the last element of a structure with more than one named member may have an incomplete array type; this is called a flexible array member. In most situations, the flexible array member is ignored. In particular, the size of the structure is as if the flexible array member were omitted except that it may have more trailing padding than the omission would imply. [...]

So if you had a s* you would allocate space for the array in addition to space required for the struct, usually you would have other members in the structure:

s *s1 = malloc( sizeof(struct s) + n*sizeof(int) ) ;

the draft standard actually has a instructive example in paragraph 17:

EXAMPLE After the declaration:

  struct s { int n; double d[]; };

the structure struct s has a flexible array member d. A typical way to use this is:

   int m = /* some value */;
   struct s *p = malloc(sizeof (struct s) + sizeof (double [m]));

and assuming that the call to malloc succeeds, the object pointed to by p behaves, for most purposes, as if p had been declared as:

    struct { int n; double d[m]; } *p;

(there are circumstances in which this equivalence is broken; in particular, the offsets of member d might not be the same).