CODEWITHSUNDEEP
c++
jsuke06
jsuke06

Reputation: 145

insert function on a linked list doesn't work

The insert function on my single linked list doesn't work. I've narrowed down the problem to the while loop statement. Since the statement has an AND operator, the entire statement either evaluates to true or false. while(value > nodePointer->value && nodePointer != nullptr) this is what I wrote, and it doesnt work

while(nodePointer != nullptr && value > nodePointer->value) I tried this and it DOES work, but I dont know why

if its all going to evaluate to true or false, why does the order matter?

class NumberList
{

private:

   struct listNode
   {
       double value;
       struct listNode *next;
   };

   listNode *head;

public:

   NumberList()
   {
       head = nullptr;
   }

   void appendNode(double);
   void insertNode(double);
   void deleteNode(double);
   void displayNode();
};

void NumberList::insertNode(double value)
{
   listNode* newNode;
   listNode* nodePointer;
   listNode* previousPointer;
   newNode = new listNode;
   newNode->value = value;
   newNode->next = nullptr;

   if (!head) //if nothing in the list
   {
       head = newNode;
       newNode->next = nullptr;
       cout << "check1" << endl;
   }

   else
   {
       nodePointer = head; //initial placement and reset
       previousPointer = nullptr;
       cout << "check2" << endl;

       while (value > nodePointer->value && nodePointer != nullptr)
       {
           previousPointer = nodePointer;
           nodePointer = nodePointer->next;
           cout << "check3" << endl;
       }

       if (previousPointer == nullptr) //if the value needs to be inserted at the front
       {
           head = newNode; //head points to newly instered node
           newNode->next = nodePointer;//newly insterted node points to the next node 
           cout << "check4" << endl;
       }
       else//everywhere else
       {

           
           previousPointer->next = newNode;
           newNode->next = nodePointer;
           cout << "check5" << endl;
       }
   }
}

Upvotes: 0

Views: 60

Answers (2)

Sam Varshavchik
Sam Varshavchik

Reputation: 118445

while (value > nodePointer->value && nodePointer != nullptr)

When nodePointer reaches the end of the road and becomes NULL, the above condition will check nodePointer->value first, before checking whether nodePointer is NULL. This is not going to end well.

You need to check for the NULL pointer as the first order of business here, and before doing anything else with it, like wondering what's the value of the object it's pointing to. The && operator evaluates the left-hand side first, and only evaluates the right hand side if it needs to. If the left-hand side is false, the right-hand side is not evaluated. That's how it works.

Upvotes: 2

john
john

Reputation: 87997

if its all going to evaluate to true or false,`

That's your mistake, it's not all going to be evaluated. In the expression A && B if A evaluates to false then B will not be evaluated at all. This is sometimes called short circuit evaluation.

Now look at your two versions when nodePointer is null.

while (value > nodePointer->value && nodePointer != nullptr)

In this case nodePointer->value causes an error, you can't dereference a null pointer. Now the other version

while (nodePointer != nullptr && value > nodePointer->value)

In this code nodePointer != nullptr is false so value > nodePointer->value is not evaluated, so it does not cause an error. That's the difference.

A similar rule applies to A || B. In this expression if A evaluates to true B will not be evaluated at all.

Upvotes: 1

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