Pointer to an array of structures

Discussion in 'C Programming' started by Frank Münnich, Jul 11, 2003.

  1. Hi there..

    My name is Frank Münnich. I've got a question about pointers that
    refer to an array of a structure.
    How do I declare that type?

    If I have declared a structure
    struct mystruc {
    int x,y,z;
    char a,b,c;

    and have furthermore declared

    mystruc data[20];

    and now I would like to have a pointer that refers to the array of
    structures, how do I do this?

    mystruc *mypointer[20] declares an array of 20 pointers, not a pointer
    referring to an array of 20 structures.

    I need this in order to pass the whole structure as a parameter in a
    function, so that the function can alter the data in the field.

    If anyone could help, it would be HIGHLY appreciated.
    Sincerely yours,

    Frank Münnich / TU Dresden
    Frank Münnich, Jul 11, 2003
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  2. Frank Münnich

    Ben Pfaff Guest

    struct x (*p)[size];
    Ben Pfaff, Jul 11, 2003
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  3. Frank Münnich

    Eric Guest

    Eric, Jul 11, 2003
  4. Frank Münnich

    Default User Guest

    This is an illegal definition. Perhaps you meant

    struct mystruc data[20];

    If you are compile this as C++, stop now. You will continue to have more
    What book are you using that doesn't explain the array to pointer

    Declare your function:

    void func (struct mystruc *funcdata);

    Then call it:

    func (data);

    Brian Rodenborn
    Default User, Jul 11, 2003
  5. That would have to be

    struct mystruc data[20];
    struct mystruc (*mypointer)[20];
    Yes, though that would be

    struct mystruc *mypointer[20];
    If you just need to modify one structure, you could have your function take
    a pointer to a structure:

    void foo (struct mystruc *mypointer) { mypointer->x = 4; }

    In fact, if you need to modify the contents of an array of structures, you
    could just pass a pointer to the first element:

    void foo (struct mystruc *mypointer, size_t array_length)
    size_t i;
    for (i = 0; i < array_length; ++i)
    mypointer.x = 4;

    Hope that helps.
    Russell Hanneken, Jul 11, 2003
  6. To all those helpers out there, THANK YOU.
    You made my day, I really appreciate your work! Thanks!!
    Sincerely yours,
    Frank Münnich
    Frank Münnich, Jul 11, 2003
  7. (Frank Münnich) wrote (11 Jul 2003) in
    / comp.lang.c:
    This is either wrong and should be
    struct mystruc data[20];
    or it is C++ and you are in the wrong place (will
    serve better)
    struct mystruc *datap;
    If you actually want explicitly "a pointer to an array of 20
    structures," use
    struct mystruc (*datap2)[20];
    But you may find member access a tiny bit trickier.
    Martin Ambuhl, Jul 12, 2003
  8. Frank Münnich

    Malcolm Guest

    The point is, you almost certainly don't. I don't think I've ever used such
    a construct in more than ten years of C programming.

    Even if you know that your array is necessarily 20 items long, almost all C
    programmers would take the address of the first element

    struct mystruct *datap = array;
    struct mystruct *datap = &array[0];

    and then use a counter to access the members of the array;

    /* set all the x members to 100 */
    datap.x = 100;
    Malcolm, Jul 12, 2003
  9. Frank Münnich

    Micah Cowan Guest

    Declaration mirrors usage. So think how you would access the root type
    of struct mystruc once you had obtained such a type.

    First you'd have to dereference the pointer:


    Then, you could access element n of the resulting array:


    NOTE: Remember that postfix operators have higher precedence than any
    others; thus, without the parens, it would be assumed that foo is an
    array of pointers, not the other way around (as you have already
    discovered). Now, you have the struct you needed! So to declare foo,
    you just use:

    struct mystruc (*data)[20];

    Note that, since declaration mirrors usage, and the postfix []
    operators bind closer than the unary *, the [] declarator
    also binds closer than the * declarator.

    Micah Cowan, Jul 14, 2003
  10. Frank Münnich


    Aug 2, 2012
    Likes Received:
    Hi, I am prati. New to this forum.I have written one C programme for structure. In this programme i passing pointer of the array of structure to the display function.
    It takes the input from the user but i am not getting proper output.I mean its display loop is not working.Please help to solve this problem.

    struct data{
    int ID;
    char name[4];
    typedef struct data record;
    void display(record *[]);
    record emp[2];
    int i;
    void main()
    record *ptr[2];
    printf("enter name");
    printf("enter ID");
    void display(record *ptr[])
    printf("%s %d",ptr->name,ptr->ID);
    prati, Aug 2, 2012
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