pointer to array c

Question

What is the following code doing   int g      9 8   int   j    g    From my understanding its creating a pointer to an array of 2 ints  But then why does this work   int x   j 0     and this not work   int x     j  0

User · Accepted Answer

The parenthesis are superfluous in your example. The pointer doesn't care whether there's an array involved - it only knows that its pointing to an int

  int g[] = {9,8};
  int (*j) = g;

could also be rewritten as

  int g[] = {9,8};
  int *j = g;

which could also be rewritten as

  int g[] = {9,8};
  int *j = &g[0];

a pointer-to-an-array would look like

  int g[] = {9,8};
  int (*j)[2] = &g;

  //Dereference 'j' and access array element zero
  int n = (*j)[0];

There's a good read on pointer declarations (and how to grok them) at this link here: http://www.codeproject.com/Articles/7042/How-to-interpret-complex-C-C-declarations

User · Answer

j 0   dereferences a pointer to int  so its type is int      j  0  has no type   j dereferences a pointer to an int  so it returns an int  and   j  0  attempts to dereference an int  It s like attempting int x   8  x 0

User · Answer

int g      9 8     This declares an object of type int 2   and initializes its elements to  9 8   int   j    g    This declares an object of type int    and initializes it with a pointer to the first element of g   The fact that the second declaration initializes j with something other than g is pretty strange  C and C   just have these weird rules about arrays  and this is one of them  Here the expression g is implicitly converted from an lvalue referring to the object g into an rvalue of type int  that points at the first element of g   This conversion happens in several places  In fact it occurs when you do g 0   The array index operator doesn t actually work on arrays  only on pointers  So the statement int x   j 0   works because g 0  happens to do that same implicit conversion that was done when j was initialized   A pointer to an array is declared like this  int   k  2     and you re exactly right about how this would be used  int x     k  0      note how  declaration follows use   i e  the syntax for declaring a variable of a type mimics the syntax for using a variable of that type    However one doesn t typically use a pointer to an array  The whole purpose of the special rules around arrays is so that you can use a pointer to an array element as though it were an array  So idiomatic C generally doesn t care that arrays and pointers aren t the same thing  and the rules prevent you from doing much of anything useful directly with arrays   for example you can t copy an array like  int g 2     1 2   int h 2   h   g      Examples   void foo int c 10       looks like we re taking an array by value     Wrong  the parameter type is  adjusted  to be int   int bar 3     1 2   foo bar      compile error due to wrong types  int 3  vs  int 10       No  compiles fine but you ll probably get undefined behavior at runtime     if you want type checking  you can pass arrays by reference  or just use std  array   void foo2 int   amp c  10       paramater type isn t  adjusted  foo2 bar      compiler error  cannot convert int 3  to int   amp   10   int baz   10      returning an array by value     No  return types are prohibited from being an array   int g 2     1 2   int h 2    g     initializing the array  No  initializing an array requires    syntax h   g     copying an array  No  assigning to arrays is prohibited     Because arrays are so inconsistent with the other types in C and C   you should just avoid them  C   has std  array that is much more consistent and you should use it when you need statically sized arrays  If you need dynamically sized arrays your first option is std  vector

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