CODEWITHSUNDEEP
carrayspointerstypes
Raja
Raja

Reputation: 6824

difference between int bar[10] vs int (*bar)[10]

int bar[10]; /* bar is array 10 of int, which means bar is a pointer to array 10 of int */

int (*bar)[10]; /* bar is a pointer to array 10 of int */

According to me they both are same, am I wrong? Please tell me.

Edit: int *bar[10] is completely different.

Thanks Raja

Upvotes: 5

Views: 702

Answers (5)

dajobe
dajobe

Reputation: 5036

There's also cdecl(1) and http://cdecl.org/

$ cdecl explain 'int bar[10]'
declare bar as array 10 of int

$ cdecl explain 'int (*bar)[10]'
declare bar as pointer to array 10 of int

Upvotes: 1

Bo Real
Bo Real

Reputation: 136

Here is a link about the C "right-left rule" I found useful when reading complex c declarations: http://ieng9.ucsd.edu/~cs30x/rt_lt.rule.html. It may also help you understand the difference between int bar[10] and int (*bar)[10].

Taken from the article:

First, symbols.  Read

     *      as "pointer to"         - always on the left side
     []     as "array of"           - always on the right side
     ()     as "function returning" - always on the right side

as you encounter them in the declaration.

STEP 1
------
Find the identifier.  This is your starting point.  Then say to yourself,
"identifier is."  You've started your declaration.

STEP 2
------
Look at the symbols on the right of the identifier.  If, say, you find "()"
there, then you know that this is the declaration for a function.  So you
would then have "identifier is function returning".  Or if you found a 
"[]" there, you would say "identifier is array of".  Continue right until
you run out of symbols *OR* hit a *right* parenthesis ")".  (If you hit a 
left parenthesis, that's the beginning of a () symbol, even if there
is stuff in between the parentheses.  More on that below.)

STEP 3
------
Look at the symbols to the left of the identifier.  If it is not one of our
symbols above (say, something like "int"), just say it.  Otherwise, translate
it into English using that table above.  Keep going left until you run out of
symbols *OR* hit a *left* parenthesis "(".  

Now repeat steps 2 and 3 until you've formed your declaration.

Upvotes: 0

Brandon E Taylor
Brandon E Taylor

Reputation: 25349

These two declarations do not declare the same type.

Your first declaration declares an array of int.

Your second declaration declares a pointer to an array of int.

Upvotes: 1

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272487

You can do this:

int a[10];

int (*bar)[10] = &a;  // bar now holds the address of a

(*bar)[0] = 5;  // Set the first element of a

But you can't do this:

int a[10];

int bar[10] = a;  // Compiler error!  Can't assign one array to another

Upvotes: 3

AnT stands with Russia
AnT stands with Russia

Reputation: 320451

They are completely different. The first one is an array. The second one is a pointer to an array.

The comment you have after the first bar declaration is absolutely incorrect. The first bar is an array of 10 ints. Period. It is not a pointer (i.e. your "which means" part makes no sense at all).

It can be expressed this way:

typedef int T[10];

Your first bar has type T, while you r second bar has type T *. You understand the difference between T and T *, do you?

Upvotes: 9

Related Questions