Thread: pointer question.... pointer theory?

what is the name of this array of 5 characters?

There is no name since the array doesn't exist. In this case 'a' is just a variable which can be assigned to the address of an array pointer. Proper usage of 'a might look like:

Code:

char (*a)[5];
char b[5];
a = &b;
*a[0] = 'a';

If you tried to assign an index value to the array 'a' points to without assigning it to an array first, you will get an error.

The parenthesis are just used to distinguish the declaration from the second example you have listed there. It is similar as to when you declare a function pointer.

so my pointer needs to be derefrenced twice? AS in this code:

Code:

#include <stdio.h>



main()
{
      char (*a)[5];
      char *h;     
      char abc[5] = "abcd";
      
      a = abc;
      
      
      printf("%c\n",a[0][2]);
      
       
      getchar();      
}

no, more like this:

Code:

int main(void)
{
      char (*a)[5];
      char *h;     
      char abc[5] = "abcd";
      
      a = &abc;
      
      printf("%c\n",(*a)[2]);
      // or
      printf("%c\n",*((*a) + 2));
       
      return 0;    
}

I think that when you put

Code:

a = &abc;

it should've been:

Code:

a = abc;

since abc is an array with 5 elements and therefore abc is the pointer constant to the first element in the array.

Right?

no. When I put

Code:

a = &abc;

I meant to put

Code:

a = &abc;

'a' doesn't point to an array. It points to a pointer which points to an array.