C ++指针，链接列表混乱

Question

I am trying to build a linked list in C++. My understanding is that the code I have created should create a node and then progressively link 4 more onto the end. Unfortunately, while I would expect to see the cout results as "12 123 1234 12345" I'm seeing "12 12 12 12" and in my main I am unable to traverse the list - it just crashes.

I have the following code:

struct listNode {

    int val;
    listNode* next;

};


int nodeCount = 0;

listNode* addToEnd(listNode* node) {

    listNode* newNode = new listNode;
    newNode->val = ++nodeCount;
    newNode->next = NULL;

    if (node == NULL) {
        return newNode;
    }

    listNode* current = node;
    cout<<"\n\n";
    do {
        if (current->next == NULL) {
            current->next = newNode;
        }
        cout<<current->val<<"\n";
        current = current->next;
    } while (current->next != NULL);
    cout<<current->val<<endl;

}


int main()
{

    listNode* first = addToEnd(NULL);

    addToEnd(first);
    addToEnd(first);
    addToEnd(first);
    addToEnd(first);

    cout<<"Third: "<<first->next->next->val;

}

Any help is appreciated, as I am at wit's end!

Answer 1

It is obvious that function addToEnd is wrong

listNode* addToEnd(listNode* node) {

    listNode* newNode = new listNode;
    newNode->val = ++nodeCount;
    newNode->next = NULL;

    if (node == NULL) {
        return newNode;
    }

    listNode* current = node;
    cout<<"\n\n";
    do {
        if (current->next == NULL) {
            current->next = newNode;
        }
        cout<<current->val<<"\n";
        current = current->next;
    } while (current->next != NULL);
    cout<<current->val<<endl;

}

Let's assume that the list already contains two nodes and consider the do-while loop inside the function. At first current_next != null so the following statement is executed

        current = current->next;

Now current points to the second node. Its data member next is equal to NULL. So the condition of the loop

    } while (current->next != NULL);

will be false and no iteration will be repeated. So we added nothing.

Also the function returns nothing if node is not equal to NULL.

Rewrite the function the following way

listNode* addToEnd( listNode* node ) 
{

    listNode* newNode = new listNode { ++nodeCount, NULL };

    if ( node == NULL) return newNode;

    listNode* current = node;

    while ( current->next != NULL ) current = current->next;

    current->next = newNode;

    return newNode;
    // or
    //return node;    
}

Take into account that this statement

cout<<"Third: "<<first->next->next->val;

outputs only the value of the third node. If you want to output all the list you should write

for ( listNode *current = first; current; current = current->next ) 
{
    std::cout << current->val << ' ';
}
std::cout << std::endl;

By the way using my function you could write in main for example the following way:)

listNode* first;

addToEnd( addToEnd( addToEnd( addToEnd( first  = addToEnd( NULL ) ) ) ) );

Answer 2

Use a for loop to get you to the last node instead of a while, and then assign the new node OUTSIDE of the loop. Trying to do it inside will result in an infinite loop (and make the code harder to read):

listNode* current;
for(current = node; current->next != NULL; current = current->next) ;
current->next = newNode;

You're also forgetting to return newNode at the end of the function.

Answer 3

You're falling off the end of a function with non- void return type. The fact that you don't use the return value does not make that ok.

6.6.3 in the Standard says that:

Flowing off the end of a function is equivalent to a return with no value; this results in undefined behavior in a value-returning function.

Answer 4

如果以if条件检查if(node==null)失败，则没有return语句。

Answer 5

Is it against the rules to use recursive functions in your question?

Why not do...

void addToEnd(listNode* node){
    if(node == NULL){
        *node = new listNode;
        node->next = NULL;
        node->val = ++nodeCount;
    }else{
        addToEnd(node->next);
    }
    return;
}

int main(){
    listNode* first = NULL;
    addToEnd(first); // 1
    addToEnd(first); // 2
    addToEnd(first); // 3
    addToEnd(first); // 4
    addToEnd(first); // Linked list is now 5 long
}

Answer 6

This is how I would have coded adding five nodes to a linked list that holds a node count. If anyone has advice it is welcome.

#include <iostream>
#include <cstdlib>

using namespace std;

struct listNode{
 int val;
 listNode* next;
};

listNode* addToEnd(listNode*, int);

int main()
{
  listNode* first = NULL;
  listNode* temp;
  int nodeCount = 1;

  for(int i = 0; i < 5; i++){
     first = addToEnd(first, nodeCount);
     nodeCount++;
  }

  temp = first;

  while(temp){
     cout << temp->val << ' ';
     temp = temp->next;
  }

  temp = first;
                               //Deallocate memory
  while(temp){                 //could do memory deallocation while displaying 
     nodeToDelete = temp;      //the value of nodeCount but wanted to illustrate  
                               //both methods individually
     temp = temp->next;

     delete nodeToDelete;   
  }  

  first = NULL;                //eliminate hanging pointer

  return 0;
}

listNode* addToEnd(listNode* node, int nodeCount) 
{

  listNode* newNode = new (nothrow) listNode;
  listNode* current = node;

  if(newNode){
     newNode->val = nodeCount;
     newNode->next = NULL;
     if (node == NULL)        
        node = newNode;       

     else{          
        while (current->next != NULL) 
           current = current->next;

           current->next = newNode;          
     }
  }
  else
     cout << "error allocationg memory" << endl;

  return node;
}