Thread: Newbie questions about pointers and linked list

  1. #1
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    Newbie questions about pointers and linked list

    Hello there,

    Suppose I've got the following structure:
    Code:
    typedef struct s_no *p_no;
    
    struct s_no {
      int value;
      p_no next;
    };
    And:
    Code:
    void insert_beginning(p_no *pL, p_no y)
    {
      y->next = *pL;
      *pL = y;
    }
    This function is used like this:
    Code:
    p_no L = NULL, x;
    
    p_no x = (p_no) malloc(sizeof(struct s_no));
    
    insert_beginning(&L, x);
    My doubt is, after calling that function, L and x will store the same address, and also L->next will be the same as x->next, right?

    So what's the difference between the function above and this one, where instead of passing &L to it I'd pass only L:

    Code:
    void insert_beginning(p_no pL, p_no y)
    {
      y->next = pL;
      pL = y;
    }
    I tried it but the results were different, L stores the 0x0 address while x stores something else, but I couldn't figure out why.

    I appreciate any help.

  2. #2
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    I don't know about others here, but I personally detest defining pointer types. I personally want to see the asterisk * whenever a variable is really a pointer; it just makes it easier to read and parse the code. I almost skipped your question because of that..

    Quote Originally Posted by koplersky View Post
    Code:
    p_no L = NULL, x;
    p_no x = (p_no) malloc(sizeof(struct s_no));
    You're declaring x twice. Typo, I presume?

    Quote Originally Posted by koplersky View Post
    My doubt is, after calling that function, L and x will store the same address, and also L->next will be the same as x->next, right?
    If L == x, and they are of the same type (p_no), then by definition L->next == x->next , yes.

    Quote Originally Posted by koplersky View Post
    So what's the difference between the function above and this one, where instead of passing &L to it I'd pass only L:
    In that case the list pL is passed by value (instead of by reference as earlier). Therefore, pL = y only affects pL in the current scope (within this function); in the caller, L will not be modified.

  3. #3
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    Thank you!

    Yes, it was typo... and about defining pointer types, the code is not mine, I was just studying it. But now you've mentioned it, I also think it's not good to define them, because using * all the time makes it easier for one to look and immediately identifying the pointer type.

    So let me see if I got it right, this line:
    Code:
    y->next = *pL;
    can modify x->next in the caller because I'm passing the node that x points to by reference, but I can't modify x itself (from the function) because I'm passing it by value. Is that correct?

    Thanks again.
    Last edited by koplersky; 10-07-2012 at 07:58 PM.

  4. #4
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    Perhaps it would be easier to read when written like this:
    Code:
    struct node {
        struct node *next;
        int       value;
    }
    
    void insert_beginning(struct node **list, struct node *one)
    {
        one->next = *list;
        *list = one;
    }
    Quote Originally Posted by koplersky View Post
    So let me see if I got it right, this line:
    Code:
    y->next = *pL;
    can modify x->next in the caller because I'm passing the node that x points to by reference, but I can't modify x itself (from the function) because I'm passing it by value. Is that correct?
    Yes, that is exactly correct!

  5. #5
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    Okay, that really cleared things up.
    Thanks a lot!

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