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];

Answer 1

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