What does this character pointer declaration mean?

Question

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C pointer to array/array of pointers disambiguation

How is char (*p)[4]; different from char *p[4];?

score 8 · Accepted Answer · answered Apr 29 '12 at 14:44

8

http://cdecl.org/

char (*p)[4]; -- declare p as pointer to array 4 of char.
char *p[4]; -- declare p as array 4 of pointer to char.

answered Apr 29 '12 at 14:44

Cat Plus Plus

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How do I use this "p as pointer to array 4 of char" – Mukul Shukla Apr 29 '12 at 14:45
Is *p[0]="Hello"; or *p[0]=(char *)malloc(10); valid for the first declaration – Mukul Shukla Apr 29 '12 at 14:46
I mean is it an array of 4 character pointers the first one? – Mukul Shukla Apr 29 '12 at 14:46
1

No, it's a pointer to array, not an array. – Cat Plus Plus Apr 29 '12 at 14:47
Is it the same of char p[4]; then p in any case will be the base address – Mukul Shukla Apr 29 '12 at 14:49

phoxis · Answer 2 · 2012-04-29T15:18:48.540

4

char (*p)[4];: p is a pointer to a char array of length 4.

                         char [4]
    points to              |
     char [4]              v

   +------+             +------+------+------+------+
   |  p   |------------>|      |      |      |      |
   +------+             +------+------+------+------+
                         char    char   char   char  

   p will point to a char [4] array. Array is not created. 
   p is intended to be assigned at address of a char [4]

char *p[4]; : p is an array of length 4, each location of the array is a pointer to char

              +------+------+------+------+
   p          |      |      |      |      |
an array      +------+------+------+------+
 itself          |      |      |      |
                 v      v      v      v  
               char*  char*  char*  char*

  p is an array and will be allocated in stack (if automatic)

edited Apr 29 '12 at 15:18

answered Apr 29 '12 at 14:53

phoxis

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for pointers, picture always helps :) – phoxis Apr 29 '12 at 15:09
2

In the case of `char (*p)[4]` it's important to note that the declaration does not create the array, only a pointer to one. – kittemon Apr 29 '12 at 15:13
@kittemon updated with the info – phoxis Apr 29 '12 at 15:19

What does this character pointer declaration mean?

2 Answers2