Using pointers in linked lists

Question

I am aware of how to use pointers in order to point to other objects such as arrays, but in the scheme of knowing how to use them in linked lists, I am confused. For Example:

 struct node {
    int info;
    struct node *next;
}
typedef struct node Node

And assume that later on in the program, dataList is set to the address of the first node of a list that holds these data in this order: 25,30,45,60,65,80,90

I am to write C statements that set ptr1 to the address of the node holding 45 and set ptr2 to the address of the node holding 80.

Given that dataList is initially the head of the list, my idea was:

  ptr1= dataList.next.next

but I am sure that is incorrect formatting or I also thought that

ptr1->next->next

might be correct, but again, I am not sure.

ptr1= dataList->next->next,Assuming dataList is the pointer to the head element. — tesseract
– tesseract, Commented Mar 4, 2014 at 6:30

Sadique · Accepted Answer · 2014-03-04 06:23:26Z

2

ptr1->next is equivalent to (*ptr1).next, i.e. it gets the member called next from the struct that ptr1 points to.

Structure dereference ("member b of object pointed to by a") a->b

answered Mar 4, 2014 at 6:23

Sadique

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Xufeng · Accepted Answer · 2014-03-04 06:28:58Z

1

As you said, dataList, ptr1 and ptr2 are pointers to struct node. So it should be ptr1 = dataList->next->next. -> is just a shortcut for a pointer to reference to the member of the struct it points to.

answered Mar 4, 2014 at 6:28

Xufeng

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Smitt · Accepted Answer · 2014-03-04 06:34:29Z

1

dataList is the one that is holding the start address of the list. So you access them all using dataList variable. Your idea of using ->next->next to point to 45 is correct just that it should be ptr1 = dataList->next->next, but we dont go about doing ->next->next->next... all over to get every element, but use a pointer to hold every next address and loop till we have got desired value.

answered Mar 4, 2014 at 6:34

Smitt

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aod · Accepted Answer · 2014-03-04 07:07:02Z

0

Doing next->next->next will work, but that isn't what we would want to achieve by using linked lists. Compare the info of each node while traversing and point ptr1 and prt2 to the node where you want to point. A sample code is given below.

PS: I am just a beginner, correct me if I am wrong.

if ( dataList != 0 ) {
        while ( dataList->next != 0)
        {
            dataList = dataList->next;
            if (dataList->info == 25 ) {
                ptr1 = dataList;
            }
            if (dataList->info == 80 ) {
                ptr2 = dataList;
            }

        }
}

answered Mar 4, 2014 at 7:07

aod

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kutichatan · Accepted Answer · 2014-03-04 07:26:33Z

0

Assume the following declaration for dataList

   struct node anode1, anode2, anode3, anode4, anode5, anode6, anode7;
   struct node *dataList = &anode1;

   anode1.info = 25;
   anode1.next = &anode2;

   anode2.info = 30;
   anode2.next = &anode3;

   anode3.info = 45;
   anode3.next = &anode4;

   anode4.info = 60;
   anode4.next = &anode5;

   anode5.info = 65;
   anode5.next = &anode6;

   anode6.info = 80;
   anode6.next = &anode7;

   anode7.info = 90;
   anode7.next = NULL;

   printf("Third node = %d\n", dataList->next->next->info); // prints 45
   printf("Sixth node = %d\n", dataList->next->next->next->next->next->info); // prints 80

answered Mar 4, 2014 at 7:26

kutichatan

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