Re: pointers to struct members
From: Barry Schwarz (schwarzb_at_deloz.net)
Date: 04/17/04
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Date: 17 Apr 2004 17:54:58 GMT
On Sat, 17 Apr 2004 14:33:34 +0200, "Emanuele Blanco"
<manub@despammed.com> wrote:
>Hi there,
>
>I just compiled a program that uses linked lists (needed it as an homework
>for my Programming course at University). It works flawlessly, even if I
>notice a thing. Here's my linked list declaration:
>
>struct node {
> float item;
> struct node *next;
>};
>
>typedef struct node *list;
>
>I need a function that removes a node in that list, without checking the
>emptyness of the list. So here's my code:
>
>void remove_key(list *lista, float key) {
> list temp=*lista;
> if (temp->item == key) {
> *lista=temp->next;
> free(temp);
> }
> else
> remove_key((list *)(temp->next), key);
>}
>
>My question is: I needed the casting to list * to make the recursive call
>work. I wasn't able to get it work using a syntax like
>remove_key(*(temp->next), key). There's a proper syntax to point to a struct
>member, or do I need the cast ? Compiling with gcc makes the incompatible
>parameter type error.
The compiler diagnostic is not the problem. It is a description of
the problem. While the cast silences the compiler, it does nothing
about the problem itself.
remove_key requires a pointer to pointer to struct as its first
argument. That means when you dereference your argument, the value
you end up with is a pointer to struct. Phrased another way, the
value in lista is an address. If you examine the appropriate number
of bytes at that address, you will find another address. It is this
second address which identifies the location of the struct.
What did your code do? You took the value of temp->next which is a
pointer to struct, that is the address of a struct. You told the
compiler: don't change the value but convert it to the form of pointer
to pointer to struct. (On most machines, the form of a pointer to
struct and pointer to pointer to struct are the same. However, the
compiler recognizes a difference between struct* and struct** so the
cast makes the compiler happy and also performs whatever magic is
necessary in case the two forms are not the same.) You then pass this
value to the next incarnation (or whatever the correct term is for the
target of the recursive call) of remove_key. remove_key knows that
the (new) value of lista points to address. Unfortunately it doesn't.
The new value of lista points to exactly what the old value of
temp->next pointed to. Namely, another instance of the struct. It
turns out that the bytes at the beginning of the struct, which
remove_key is trying to interpret as an address, are actually part of
the float that is the first item in the struct. Welcome to undefined
behavior.
The moral of the story is you cannot lie to your compiler. When you
use a cast, what you are really saying to the compiler is that even if
this expression doesn't look like this new type, it really is.
Therefore, don't change the value but do "reformat" it so it "looks
like" the type specified. If this is not the situation in your code,
then the cast is the wrong solution.
remove_key requires a value (C always passes by value) that points to
a pointer. The proper method of constructing such a value when all
you have is the original pointer is the "address of" operator, the &.
What your code needs is
remove_key(&(temp->next), key);
Since -> has higher precedence than &, the internal parentheses are
superfluous but some like to keep them for the visual hint.
The next question is: why are you doing this recursively? It is
relatively simple (and much easier to debug) to do this with a loop.
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