I'm pretty sure the code is undefined once you do x+1 (that is, treat an array l...

mikeash · on Oct 18, 2011

From the near-final draft of the C99 standard I happen to have sitting around:

"Except when it is the operand of the sizeof operator or the unary & operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue. If the array object has register storage class, the behavior is undefined."

In short, the moment you write a variable of array type and it's not one of those very limited cases, it immediately becomes a pointer to the first element. x+1 is completely valid.

TeMPOraL · on Oct 18, 2011

What about

  *(x+1)

that is equivalent to x[1]? Here, in principle, you do x+1 before applying *.

EDIT

HN formatting magic.

mtoddh · on Oct 18, 2011

Fun tidbit - this equivalence also means you can write it as 1[x]:

  #include <stdio.h>
  int main()
  {
    int x[5] = {1,2,3,4,5};
    printf("%d, %d\n", x[1],1[x]);
    return 0;
  }