"array"

  Below, an array is allocated dynamically.

#include <stdio.h>
#include <stdlib.h>

int main( void )
{ char *array_pointer = malloc( 10 * sizeof *array_pointer );
  if( !array_pointer )return EXIT_FAILURE;
  *array_pointer = 'a';
  free( array_pointer ); }

  But is it really an array according to the C spec?

  C only defines "array type", not array, but it uses ISO/IEC 2382:2015
  as a normative reference, and ISO/IEC 2382:2015 says:

|array
|An aggregate that is an instance of an array type and each
|element or appropriate group of elements in which may be
|referenced randomly and independently of the others.
ISO/IEC 2382-15 (1998), 15.03.08 (Can't access newer versions!)

  . That block of memory that malloc allocated, does it have an
  array type? I'm not sure. So are we allowed to call it an array?
  I think the C community /would/ mostyly call this an array.

  Here's an attempt to give it an explicit array type:

#include <stdio.h>
#include <stdlib.h>

int main( void )
{ char( *array_pointer )[ 10 ]= malloc( sizeof *array_pointer );
  if( !array_pointer )return EXIT_FAILURE;
  ( *array_pointer )[ 0 ]= 'a';
  free( array_pointer ); }

  Is it now more of an array than before?

  In n3467:

|The effective type of an object that is not a byte array, for
|an access to its stored value, is the declared type of the
|object. 83)
from 6.5p6

|83) Allocated objects have no declared type.
footnote for 6.5p6

|For all other accesses to a byte array, the effective type of
|the object is simply the type of the lvalue used for the access.
from 6.5p6

  In my example programs, the lvalue for the access has the
  type "int". So, I can see that there's an int object, and
  - in a similar way - that there are other int objects behind it,
  but not necessarily that they form an array.