pointer probs. with structs.

This is a discussion on pointer probs. with structs. within the C Programming forums, part of the General Programming Boards category; Hi, I'm having problems compiling the following simple code. I'm trying to see how this works on a small scale ...

  1. #1
    Registered User dinjas's Avatar
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    pointer probs. with structs.

    Hi, I'm having problems compiling the following simple code. I'm trying to see how this works on a small scale before I move the idea into this program I'm working on. Quzah kind of explained this to me the other day but it's not quite clicking for this exact application...

    Code:
     #include <stdio.h>
         
    struct node
    {
      int data;
      struct node* next;
    };
    
    void foo(struct node** n )
    {
      *n->data = 5;
    }
     
    int main( void )
    {
    
      struct node* root = NULL;
      root = malloc(sizeof(struct node));
      root->data=3;
      printf("root->data = %d", root->data);
      foo( &root );
      printf("root->data = %d", root->data);
      return 0;
    
    }
    My compiler is telling me that in my function foo, data is not a member of the struct. I'm sure this has to do with the dereferencing but it doesn't seem to matter how many *'s I put, 0, 1, or 2. I still get the message. I pass in a ptr to a ptr. as an address of a ptr. But to access the data, I use the ->data which is the same as using one * for dereferencing right? So, I need to use 1 more to change the value of the data? Also, the compiler doesn't like my malloc call in main. Says I'm making a pointer from an integer without a cast. I've used malloc like this before though, so ??? I'm I doin bad stuff?
    I need to be able to pass pointers to structs to functions and be able to modify elements in the structures and have the changes stay. Thanks
    dinjas
    straight off the heap

  2. #2
    Registered User hk_mp5kpdw's Avatar
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    Code:
    (*n)->data = 5;
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
    -Christopher Hitchens

  3. #3
    Registered User dinjas's Avatar
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    ughhhh
    ( 1 hour
    ) 2 hour, 3 hour, 4 ...

    so any time you dereference something with the -> after it you need the parens? never would have thought of it..

    thanks!
    straight off the heap

  4. #4
    Super Moderator
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    You'll find a precedence table useful for this type of thing. It'll help you see how an expression will be interpreted.

  5. #5
    Registered User hk_mp5kpdw's Avatar
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    For your purpose here, changing the value of the struct member, you don't need to make the function accept a pointer to a pointer to struct node. A single pointer to type struct node would suffice. This would eliminate the need for passing the address of root to the foo function and the extra dereferencing and use of parenthesis steps within the foo function. You could have simply done this and achieved the same effect:

    Code:
    struct node
    {
      int data;
      struct node* next;
    };
    
    void foo(struct node* n )  // Single pointer
    {
       n->data = 5;  // No extra parenthesis and dereferencing needed here
    }
     
    int main( void )
    {
    
      struct node* root = NULL;
      root = malloc(sizeof(struct node));
      root->data=3;
      printf("root->data = %d", root->data);
      foo( root );  // No address of operator (&) needed
      printf("root->data = %d", root->data);
      return 0;
    
    }
    Now, if on the other hand you wished to change what root pointed to by passing it in as a parameter, then you would need to use a pointer to a pointer to struct node, i.e.:

    Code:
    struct node
    {
        int data;
        struct node* next;
    };
    
    void foo1(struct node* n )
    {
        n = (node*) malloc(sizeof(struct node));
        n->data = 3;
    }
    
    void foo2(struct node** n )
    {
        *n = (node*) malloc(sizeof(struct node));
        (*n)->data = 10;
    }
     
    int main( void )
    {
    
        struct node* root = NULL;
    
        foo1( root);
        if( root != NULL )
            printf("root->data = %d after foo1\n", root->data);
        else
            printf("root is still NULL after foo1\n");
    
        foo2( &root );
        if( root != NULL )
            printf("root->data = %d after foo2\n", root->data);
        else
            printf("root is still NULL after foo2\n");
    
        return 0;
    }
    Which outputs:
    Code:
    root is still NULL after foo1
    root->data = 10 after foo2
    Or, foo2 could have been done as a void function returning a pointer to struct node and having the same effect:
    Code:
    struct node* foo2()
    {
        struct node *n = malloc(sizeof(struct node));
        n->data = 10;
        return n;
    }
    Which would be called like so:
    Code:
    struct node *root = NULL;
    ...
    root = foo2();
    if( root != NULL )
        printf("root->data = %d after foo2\n", root->data);
    else
        printf("root is still NULL after foo2\n");
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
    -Christopher Hitchens

  6. #6
    Registered User dinjas's Avatar
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    Thanks for the clarification. Is one method better practice than the other?
    i.e.

    Code:
    foo2(&root);
    vs.

    Code:
    root=foo2();
    Does it just depend on the implementation?
    (trying not to develop too many bad habits)

    thanks again.
    straight off the heap

  7. #7
    ATH0 quzah's Avatar
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    If you need to pass a value and then modify it inside the function, having the change persist outside of the function, a pointer is the way to go. If you don't need to pass anything, but are simply getting a value from a function and assigning it to something outside the function, then return value assignment works well.

    Really it's a matter of preference. But the above is a good general rule of thumb.

    Quzah.
    Hope is the first step on the road to disappointment.

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