Are
int (*x)[10];
and
int x[10];
equivalent?
According to the "Clockwise Spiral" rule, they parse to different C declarations.
For the click-weary:
The ``Clockwise/Spiral Rule'' By David Anderson
There is a technique known as the ``Clockwise/Spiral Rule'' which enables any C programmer to parse in their head any C declaration!
There are three simple steps to follow:
1. Starting with the unknown element, move in a spiral/clockwise direction;
when ecountering the following elements replace them with the
corresponding english statements:
[X] or []
=> Array X size of... or Array undefined size of...
(type1, type2)
=> function passing type1 and type2 returning...
*
=> pointer(s) to...
2. Keep doing this in a spiral/clockwise direction until all tokens have been covered.
3. Always resolve anything in parenthesis first!
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.
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 int
s
int x[10];
x
is an array of 10 int
s
int *x[10];
x
is an array of 10 pointers to int
s
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.
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.
No. First one declares an array of 10 int pointers and second one declares an array of 10 ints.
来源:https://stackoverflow.com/questions/5436350/equivalent-c-declarations