Help with linked lists/ pointers

**anything25** · 06-26-2009

I am trying to write a simple C program that creates a linked list and then prints out the values stored in the list. I am not sure what part of the code isn't correct but when it prints out the entries it will only print out one entry and it looks to be a memory location instead of the value stored in the node. Also, and this is the part that really throws me off, sometimes when i compile the code i get the error "invalid conversion from 'void*' to 'node*' " on every line that I use malloc and other times it compiles with no errors. Here is the code:

Code:

#include <stdio.h>
#include <stdlib.h>

typedef struct node *nodePtr;

struct node {
  int value;
  nodePtr next;
};

typedef struct node node;

void createNewNode(nodePtr head, nodePtr newNode);
void printList(nodePtr head);

int main(int argc, const char * argv[]) {
    nodePtr first = malloc(sizeof(node));
    nodePtr newNode = malloc(sizeof(node));
    int numberOfEntries = 0;

    first->next = NULL;

    printf("Number of Entries: ");
    scanf("%d", &numberOfEntries);
    printf("\n");

    if(numberOfEntries == 1) {
        printf("Enter Value: ");
        scanf("%d", first->value);
        printf("\n");
    } else {
        int i = 0;
        while(i < numberOfEntries) {
            printf("Enter Value: ");
            scanf("%d", first->value);
            printf("\n");

            createNewNode(first, newNode);
            i++;
        }
    }
    printList(first);

    return 0;
}

void createNewNode(nodePtr head, nodePtr newNode) {
    newNode->next = head;
    head = newNode;
}

void printList(nodePtr head) {
    nodePtr current = malloc(sizeof(node));
    current = head;

    while(current->next != NULL) {
        printf("Entry: %d\n", current->value);
        current = current->next;
    }

    printf("Entry: %d\n", current->value);

    free(current);
}

If someone could take a look at this and point me to where I've made a mistake I would greatly appreciate it.

**Zlatko** · 06-26-2009

You're passing head to createNewNode by value.

**MK27** · 06-26-2009

You are confusing yourself with this:

Code:

typedef struct node *nodePtr;

struct node {
  int value;
  nodePtr next;
};

Just use:

Code:

typedef struct _node {
  int value;
  struct _node *next;
} node;

Now you can properly refer to a node, and a *node.

All linked list functions should be passed pointers:

Code:

void createNewNode(node *head, node *newNode);
void printList(node *head);

**anything25** · 06-26-2009

I looked at createNewNode and tried changing a couple things, but that only seemed to make it worse. I am not sure that I completely understand what you mean by pass head in by value (I am pretty new to programming). Also, while that I don't doubt that the function is flawed, it still doesn't explain why the code would not work correctly if there is only one element in the list since that function would never be executed.

**MK27** · 06-26-2009

Originally Posted by anything25

Also, while that I don't doubt that the function is flawed, it still doesn't explain why the code would not work correctly if there is only one element in the list since that function would never be executed.

It is very important that you use one of the forms recommended by either me or laserlight. You need to pass pointers without obfuscating the fact; this will make debugging your program that much easier as well.

You are not actually passing by value, it just looks that way because you have created a datatype which is a pointer. There is no need for that and it is obviously confusing you *and* the people you are looking to for help.

**Zlatko** · 06-26-2009

Originally Posted by anything25

I looked at createNewNode and tried changing a couple things, but that only seemed to make it worse. I am not sure that I completely understand what you mean by pass head in by value (I am pretty new to programming). Also, while that I don't doubt that the function is flawed, it still doesn't explain why the code would not work correctly if there is only one element in the list since that function would never be executed.

OK, there may be a number of issues here, but remember if you have a function that takes an integer

Code:

void foo(int i)
{
     i = i + 1
}
int main()
{
   int i  = 0;
   foo(i);
}

and you increment i in the function, the caller will not see the change because i is a copy of what the caller passed in.

Now you have a pointer called first in main and you're passing it into creatNewNode as head and you want first in main to be changed to point to newNode so you must pass in the address of the variable to be changed.

The function should be

Code:

void createNewNode(nodePtr* head, nodePtr newNode)
{
   newNode->next = *head;
   *head = newNode;
}

May I ask, what development tools are you using?

Best regards
Zlatko

**MK27** · 06-26-2009

Originally Posted by Zlatko

and you increment i in the function, the caller will not see the change because i is a copy of what the caller passed in.

This is a good explanation of the difference between pass by value/pass by reference *but* Zlatko, look, the OP is passing by reference:

Code:

typedef struct node *nodePtr;

That is the first thing which needs to be fixed tho, since this is not a nice way to program at all.

**anything25** · 06-26-2009

Thanks for all of the replies. I see what you guys mean about the form it is in and have changed the code so that it shouldn't be as confusing. Here is the changed code:

Code:

#include <stdio.h>
#include <stdlib.h>

typedef struct node {
  int value;
  struct node *next;
} node;

void createNewNode(node *head, node *newNode);
void printList(node *head);

int main(int argc, const char * argv[]) {
    node *first;
    first = malloc(sizeof(node));
    node *newNode;
    newNode = malloc(sizeof(node));
    int numberOfEntries = 0;

    first->next = NULL;

    printf("Number of Entries: ");
    scanf("%d", &numberOfEntries);
    printf("\n");

    if(numberOfEntries == 1) {
        printf("Enter Value: ");
        scanf("%d", first->value);
        printf("\n");
    } else {
        int i = 0;
        while(i < numberOfEntries) {
            printf("Enter Value: ");
            scanf("%d", first->value);
            printf("\n");

            createNewNode(first, newNode);
            i++;
        }
    }
    printList(first);

    return 0;
}

void createNewNode(node *head, node *newNode) {
    newNode->next = head;
    head = newNode;
}

void printList(node *head) {
    node *current;
    current = malloc(sizeof(node));
    current = head;

    while(current->next != NULL) {
        printf("Entry: %d\n", current->value);
        current = current->next;
    }

    printf("Entry: %d\n", current->value);

    free(current);
}

I am still pretty lost with what else is wrong with the code.... Linked lists are confusing the hell out of me lol

**tabstop** · 06-26-2009

Originally Posted by MK27

This is a good explanation of the difference between pass by value/pass by reference *but* Zlatko, look, the OP is passing by reference:

Code:

typedef struct node *nodePtr;

That is the first thing which needs to be fixed tho, since this is not a nice way to program at all.

I think you've lost the plot, MK. No matter what the typedef is, this code:

Code:

void createNewNode(nodePtr head, nodePtr newNode) {
    newNode->next = head;
    head = newNode;
}

cannot possibly change head in the main function. The fact that it is a pointer is completely irrelevant.

**laserlight** · 06-26-2009

Originally Posted by MK27

Just use:

Alternatively, you could shift that later typedef to before the struct definition:

Code:

typedef struct node node;

struct node {
  int value;
  node *next;
};

**Zlatko** · 06-26-2009

Keep at it until you understand it.
Step through it with a debugger.
Ask questions
We want you to pass your course!

EDIT:
IMHO there was nothing wrong with your originaly typedef-ing

**anything25** · 06-26-2009

Thank you everyone so much for the help! I've got the code working and also understand pointers a whole lot more then when I started. Thanks again!

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