Equivalent C declarations

Question

Are

int (*x)[10];

and

int x[10];

equivalent?

According to the \"Clockwise Spiral\" rule, they parse

Answer 1

They are not equal. in the first case x is a pointer to an array of 10 integers, in the second case x is an array of 10 integers.

The two types are different. You can see they're not the same thing by checking sizeof in the two cases.

Answer 2

For me, it's easier to remember the rule as absent any explicit grouping, () and [] bind before *. Thus, for a declaration like

T *a[N];

the [] bind before the *, so a is an N-element array of pointer. Breaking it down in steps:

   a     -- a
   a[N]  -- is an N-element array
  *a[N]  -- of pointer
T *a[N]  -- to T.

For a declaration like

T (*a)[N];

the parens force the * to bind before the [], so

    a      -- a
  (*a)     -- is a pointer
  (*a)[N]  -- to an N-element array
T (*a)[N]  -- of T

It's still the clockwise/spiral rule, just expressed in a more compact manner.

Answer 3

I tend to follow The Precedence Rule for Understanding C Declarations which is given very nicely in the book Expert C Programming - Deep C Secrets by Peter van der Linden

A - Declarations are read by starting with the name and then reading in 
precedence order.

B - The precedence, from high to low, is:
        B.1 parentheses grouping together parts of a declaration
        B.2 the postfix operators:
        parentheses () indicating a function, and
        square brackets [] indicating an array.
        B.3 the prefix operator: the asterisk denoting "pointer to".

C If a const and/or volatile keyword is next to a type specifier (e.g. int, 
        long, etc.) it applies to the type specifier. 
        Otherwise the const and/or volatile keyword 
        applies to the pointer asterisk on its immediate left.

Answer 4

No. First one declares an array of 10 int pointers and second one declares an array of 10 ints.

Answer 5

Follow this simple process when reading declarations:

Start at the variable name (or innermost construct if no identifier is present. Look right without jumping over a right parenthesis; say what you see. Look left again without jumping over a parenthesis; say what you see. Jump out a level of parentheses if any. Look right; say what you see. Look left; say what you see. Continue in this manner until you say the variable type or return type.

So:

int (*x)[10];

x is a pointer to an array of 10 ints

int x[10];

x is an array of 10 ints

int *x[10];

x is an array of 10 pointers to ints