get and set value from the same function with the same params

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Hello everyone,
I am making a linked list code library,
i 've designed a function that return the value of the element at any in range given index
now i want to know how to make this function set and get without actually passing the new value as a parameter in the function

here 'is the function

C++

int32 list_element_at(llist list, int32 n)
{
   struct node *temp;
   int32 value;
   
   temp = list.head;
   while(temp != null)
   {
      if(temp -> index != n)
      {
         temp = temp -> next;
         continue;
      }
      value = temp -> data;
      break;
   }
   
   return value;
}

i want to use it like this;

C++

a = list_element_at(listo, 2);(allready done)

and

C++

list_element_at(listo , 2 ) = 50;
printf(list_element_at(listo, 2));

----> 50

Thanks in advance,
z3ngew

Posted 6-Sep-13 22:05pm

z3ngew

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OriginalGriff · Accepted Answer · 2013-09-06T22:49:00

You can't, not without changes to your function - it returns an integer - the value of the node - rather than the node itself, so the code fragment you show:

C++

list_element_at(listo , 2 ) = 50;

Is the equivalent of trying to say:

C++

17 = 50;

Which no modern compiler will allow you to do, for obvious reasons. :laugh:

You can do it, if you make your function return the node, instead of the value of the node:

C++

node* list_element_at(llist list, int32 n)
   {
   struct node *temp;
   temp = list.head;
   while(temp != null)
      {
      if(temp->index == n)
         {
         break;
         }
      temp = temp->next;
      }
   return temp;
   }

(I've taken the liberty of "tightening up" your code a little as well, just to make it a bit more readable)
You can then set the value of the node directly:

C++

list_element_at(listo, 2)->data = 50;

But you will have to watch out for invalid indexes!

nv3 · Accepted Answer · 2013-09-06T22:51:00

The solution is simple: You just have to return a reference to data member:

C++

int32& list_element_at(llist list, int32 n)
{
   ...
   return temp->data;

Just a single ampersand and you are done :-). Now, what that causes behind the scenes is that the compiler will pass a pointer to the data cell instead of a value as the return value of your function. But as you have marked it as a reference, the compiler remembers that it has to dereference this pointer before using it. In other words, this is like writing:

C++

int32* list_element_at(llist list, int32 n)
{
   ...
   return &temp->data;
}

...
*(list_element_at (listo , 2)) = 50;

and the compiler does away with the two asterisks automatically.

While we are at it, I suggest that you give your variables more meaningful names. "temp" is for example no good choice. How about writing your little function like this:

C++

int32 list_element_at (llist list, int32 index)
{
   struct node *pNode;
   
   pNode = list.head;
   while (pNode )
   {
      if (pNode->index == index)
          return pNode->value;
   }
   return INT_MIN; // this is still a problem!
}

In your current version you return an undefined value if you can't find an element with the given index. But what should your function return instead? You have to find a value that is different from all the values in your nodes. I have chosen INT_MIN, which is the lowest integer value that can be represented in an int. But still this is no good solution. In the version that returns a reference, the problem is even bigger, as you have to return something.

My suggested solution is not to return a value but a node pointer. And in case you can't find the node you would return a null pointer:

C++

struct node* list_element_at (llist list, int32 index)
{
   struct node *pNode;

   pNode = list.head;
   while (pNode )
   {
      if (pNode->index == index)
          return pNode;
   }
   return 0; // no element found
}

That also solves your problem with assigning to the element. You can now write:

C++

struct node* pNode = list_element_at (listo , 2);
if (pNode)
    pNode->data = 50;

That looks a lot better.
<END OF LESSEN MODE>

get and set value from the same function with the same params

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