Thread: array of struct pointers

  1. #1
    Registered User
    Join Date
    Aug 2008

    array of struct pointers

    hi there,

    i am trying to create an array of struct pointers, set each element in the array to null, and (later) then assign struct values to certain positions within the array... and i am getting confused with accessing the
    elements in the array ...

    if i have an array of char pointers...

             char *charPtr = malloc(sizeof(char*)*5);
    and go

      	  charPtr[2] = 'b';
      	  printf(" the char pointer is :%c: \n",charPtr[2]);
    i get

              the char pointer is :b:
    but if i go...

    typedef struct indexType
    	nodePtr head;
    	nodePtr tail;
           int count;
    } IndexType;
    typedef IndexType * IndexTypePtr;
    int main()
           IndexTypePtr tester = NULL;
           tester = malloc(sizeof(IndexTypePtr)*10);
           tester[2] = NULL;
      return 0;

    I get an assignment type mismatch ?

            assignment type mismatch:
            struct indexType {pointer to struct node {..} head, pointer to struct node {..} tail,   
            int count} "=" int
    is this because i am trying to assign null to something that has elements like head, tail

    but if that is true then how does

             IndexTypePtr  tester = NULL;
    work ??


  2. #2
    Registered User
    Join Date
    Sep 2007
    char *charPtr = malloc(sizeof(char*)*5);
    OK, there's a problem here. You say you're making an array of char pointers (and we'll pretend that a pointer is an array for the sake of this discussion). But charPtr is, in that case, an array of char, not an array of pointers. As such, you should not be using sizeof(char *) to size the elements, but sizeof(char). Of course, sizeof(char) is 1 so it's redundant to multiply by it, but it's OK for illustration.

    What you really want for an array of char pointers is something like:
    char **a = malloc(sizeof *a * 5);
    You can use sizeof(char *) there if you want, but I prefer not scattering type names around. Now you have an array of char pointers, and you can do this sort of thing:
    a[0] = NULL;
    a[1] = "string";
    /* a[1][0] is 's' now */
    The same holds for your array of struct pointers. The way you have it is an array of structs, and so tester[2] is a struct, not a pointer. Again, you want a pointer to a pointer:
    struct foo **a = malloc(sizeof *a * 5); /* Typedefing a pointer makes things unclear, in my opinion */
    a[0] = NULL; /* perfectly OK now */
    The thing with this, though, is that for each element you have to point it somewhere useful. That could be via a malloc():
    a[0] = malloc(sizeof *a[0]);
    a[0]->head = whatever;

  3. #3
    Registered User
    Join Date
    Apr 2009
    sorry to interrupt but whats a node?

  4. #4
    Kernel hacker
    Join Date
    Jul 2007
    Farncombe, Surrey, England
    It comes from the word "knot" in Latin. In this case, I'd say that it is a "knot" or "junction" in a linked list structure (or similar).

    Linked list - Wikipedia, the free encyclopedia

    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  5. #5
    Registered User
    Join Date
    Aug 2008
    thanks cas, very helpful response

    i got confused there for a while.... ;p

    notation is exactly what i was trying to do and now it works great

    very cool i can now add any sort of data structure ,bst,linked lists etc... from this
    (index / array not really sure what to call it ) ... which i will create a hash for the relative positions ... exciting ..

    and yeah node here is part of a bst or list..

    thanks again to all of the response

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