A basic linked list question

This is a discussion on A basic linked list question within the C Programming forums, part of the General Programming Boards category; Hey, Im trying to delete a specified node from a linked list. I've inlcuded my whole program but the problem ...

  1. #1
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    A basic linked list question

    Hey, Im trying to delete a specified node from a linked list.
    I've inlcuded my whole program but the problem is in removeNode() function.
    The program never enters the if statement in removeNode()

    The function removeNode is not finished. It only checks for the special case where the node to be removed is the head of the linked list.

    What is wrong with my if statement ?
    Thanks


    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    /*linked list*/
    struct node {
    	int data;
    	struct node* next;
    };
    
    /*prototypes*/
    void push(struct node**, struct node*);
    void printList(struct node*);
    struct node* createNode(int);
    int removeNode(struct node**, struct node*);/*remove the specified link*/
    
    
    int main(void){
    	struct node* head = NULL;
    
            push(&head, createNode(5));
    	push(&head, createNode(20));
    	printList(head);
    	removeNode(&head, createNode(20));
    	printList(head);
    
    	return 0;
    }
    void push(struct node** head, struct node* newNode){
    	struct node* current = *head;
    
    	newNode->next = *head;
    
    	*head = newNode;
    }
    void printList(struct node* head){
    	struct node* current = head;
    	while(current != NULL){
    		printf("%d ", current->data);
    		current = current->next;
    	}
    	printf("\n");
    }
    struct node* createNode(int data){
    	struct node* newNode = malloc(sizeof(struct node));
    
    	newNode->data = data;
    	newNode->next = NULL;
    
    	return newNode;
    }
    int removeNode(struct node** head, struct node* node){
    	struct node* current = *head;
    	if(current == node){
    		*head = current->next;
    		return 1;
    	}
    	return 0;
    }

  2. #2
    and the Hat of Guessing tabstop's Avatar
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    You have two different nodes with value 20 -- the one that's in the list from the push statement, and the one you pass to your deletenode function. Since they're not the same object,....

  3. #3
    Registered User claudiu's Avatar
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    That is not how you remove a node from the list. There is no recursion nor any iteration in that function so how are you actually traversing the list?
    1. Get rid of gets(). Never ever ever use it again. Replace it with fgets() and use that instead.
    2. Get rid of void main and replace it with int main(void) and return 0 at the end of the function.
    3. Get rid of conio.h and other antiquated DOS crap headers.
    4. Don't cast the return value of malloc, even if you always always always make sure that stdlib.h is included.

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    when comparing two linked list elements, what exactly is compared?
    The addresses?

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    Quote Originally Posted by claudiu View Post
    That is not how you remove a node from the list. There is no recursion nor any iteration in that function so how are you actually traversing the list?
    Like i said in my post. I am working on the special case where the node to be removed is the head of the list. I am not finished with it yet. I just got stuck on how to compare nodes so I figured I'd better figure this out first before I take care the more general cases.

  6. #6
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by gp364481 View Post
    when comparing two linked list elements, what exactly is compared?
    The addresses?
    Right now that's what you're comparing. Given that you are passing in an actual object, that would be the obvious thing to compare -- is the object in the list the actual object that was passed in, rather than "some other object that happens to have the same value/name/other characteristic".

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    Quote Originally Posted by tabstop View Post
    Right now that's what you're comparing. Given that you are passing in an actual object, that would be the obvious thing to compare -- is the object in the list the actual object that was passed in, rather than "some other object that happens to have the same value/name/other characteristic".
    See, that was my problem. I was passing in a node with identical data but it wasnt the actual/same object as the one inside the linked list. I've now realized that it has to be the same object and that I cannot compare nodes in a linked list as I would compare primitive types.
    Thank you

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    Gp, I disagree with you.

    Pointers ARE primitive data types. They are the MOST primitive data types, arguably.

    In this line of code, what are you passing to the function?
    Code:
    removeNode(&head, createNode(20));
    head is a pointer - and pointers hold values that are addresses. So you're passing the address of the address, in that first parameter I bolded, above.

    Make sense?

  9. #9
    Algorithm Dissector iMalc's Avatar
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    I would never use a variable with the same name as a type in C. It sure as heck wont let you do that in C++, and as far as C goes, I've never wanted to make that deliberate mistake, though from what I'm seeing here it actually does compile? - yuck!
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  10. #10
    ATH0 quzah's Avatar
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    It won't let you do that in C either. You can't make a variable called 'int'. But, 'node' is not the same thing as the type 'struct node', which is why it lets you do it.


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  11. #11
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Adak
    Pointers ARE primitive data types. They are the MOST primitive data types, arguably.
    Technically, the C standard lists pointers among the derived types, but I don't think gp364481 had this concept of derived types in mind.
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