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Array of pointers question

  1. Oct 20, 2008 #1
    in this code i got a chat type pointer which is an array
    who consists out of 4 char cells

    in order that each pointer cell of this array will point to some place
    we need to give it an address some thing like &chr

    "Monday morning" is not an address and its not a single char

    this misunderstanding is for every value in this array

    and we have 4 cells and in char arrays the last cell has to be \0
    so there only place for 3 values

    the 4th cell is www.iota-six.co.uk instaed of \0

    but its wrong

    every char array must end with \0
    and here we changed \0 to www.iota-six.co.uk

    this is wrong

    Code (Text):

    char *strings[4] = { "Monday morning" , "6AM" ,"Sunrise" , "www.iota-six.co.uk"};
  2. jcsd
  3. Oct 20, 2008 #2

    D H

    Staff: Mentor

    What you declared is a valid (but better as a const char * array), useful construct that is used quite often.

    You explanation is not what is happening. To begin, char *strings[4] declares an array of 4 "pointers to char", not an indeterminately sized array of character cells of 4 characters each. Big difference. The latter is char strings[][4];.

    You don't want a single character with the above declaration. What you want is a set of things that are can be interpreted as a pointer to a character -- and that is exactly what "Monday morning" is.

    It looks like you might be taking too big a bite of the C language. You first should understand why this works:
    Code (Text):

       const char * string = "Monday morning";
       printf ("As of my post time, it is ridiculously early on %s\n", string);
       string = "Monday afternoon";
       printf ("But if I am lucky I might be able to sneak in a nap on %s", string);
  4. Oct 20, 2008 #3


    User Avatar

    Code (Text):

    char *strings[4] = { "Monday morning" , "6AM" ,"Sunrise" , "www.iota-six.co.uk"};
    What you get is Four pointers. Your declaration is like below:

    char * strings[0]= "Monday morning";
    char * strings[1]= "6AM";
    char * strings[2]= "Sunrise";
    char * strings[3]= "www.iota-six.co.uk"

    NOTE: I write these code above to demonstrate a "pointer array" definition. You can't compile it in the real code. Each pointer has point to a char array with a '\0' added at the end implicitly by the compiler.
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