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Help on pointers

Discussion in 'C' started by debabrata, Nov 4, 2010.

  1. debabrata

    debabrata New Member

    Hi, being an student of M.C.A 1st sem ,I personally request all the members to help me on the said tropic.
    THANK YOU ALL.:)
     
  2. xpi0t0s

    xpi0t0s Mentor

    Code:
    int x=5;
    int *ptr_to_x = &x;
    printf("x is %d\n", *ptr_to_x);
    
    Does that help? If not, what are your specific questions?
     
    debabrata likes this.
  3. debabrata

    debabrata New Member

    Thanks,xpi0t0s.
    Actually I need some help in accessing structures using pointers.
    For example:- write a program to access structures , by creating a set of structures using Dynamic Memory Allocation
     
  4. xpi0t0s

    xpi0t0s Mentor

    Ok, so where exactly are you stuck? Do you know how to define a structure? To allocate memory dynamically? Make a start and see how far you get; you might surprise yourself and be able to do it all on your own. The chances are that you've already got the information you need.
     
  5. debabrata

    debabrata New Member

    Ya I know how to define a structure & allocate memory dynamically, but I can't understand how to access those structures.

    For ex:-

    struct sname *p;
    p=(struct sname *)malloc(n*(sizeof(struct sname)));

    now I have allocated memory for 'n' structures. Now kindly help me in using them.
     
  6. xpi0t0s

    xpi0t0s Mentor

    It's very simple. Suppose you had an automatic version:
    struct sname M;
    and let's say that struct sname contains an integer j

    you would access this as M.j, and assign to it with, e.g., M.j=10. I assume you already know this.

    With the code you've given above, array semantics will work, so p[5] is the 6th element of p, which will point to p+5*sizeof(struct sname) (in bytes). The compiler knows that p is a pointer to type struct sname so operations like p++ will use the size of the structure, so if p=1000 and sizeof(struct sname) is 10, then p++ will set p to 1010, not to 1001.

    [[ So this is why I added "(in bytes)", because if you actually do p+5*sizeof(struct sname) then this will evaluate to 1000+(5*10)*10=1500. ]]

    So to access the structure members just replace M with p[5]. Assign a value with p[5].j=10;, just as in the automatic version. Copy this to another variable with, e.g. int x=p[5].j;
     
  7. debabrata

    debabrata New Member

    Does '.' operator works ,when we are using addresses or we have to use the '->' operator, which one is better?
     
  8. xpi0t0s

    xpi0t0s Mentor

    It's not a case of one being better than the other; these are different syntaxes for different situations. . accesses the members of an automatic/static structure, and -> accesses members via a pointer. So:
    struct sname A, *B;

    defines A as a struct sname and B as a pointer to a struct sname, to access member j in each, you need A.j or B->j. (Note however that B doesn't point to anything at present, so B->j is going to give you undefined behaviour.)

    A->j won't work because A isn't a pointer, and B.j won't work because B _is_ a pointer. Both these will throw compiler errors.

    B->j is effectively a shortcut of, and neater than, (*B).j.
     

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