Two methods of swapping integers using pointers, which is better?[C]

Question

i have encountered two methods of swapping two integer values and I’m confused about the difference.

Method1(temp is a pointer):

 void Swap(int *a, int *b)
 {
  int *temp = a; 
  *a = *b;
  *b = *temp;
  } 

Method2(temp is not a pointer):

  void Swap(int *a, int *b)
  {
  int temp = *a; 
  *a = *b;
  *b = temp;
   }

I don’t understand the difference between two methods, and is one method better than the other?

Answer 1

Perhaps it's easier to understand the first (wrong) function if we draw it out?

When the function is called, it will be something like this:

+---+      +-------------+
| a | ---> | value for a |
+---+      +-------------+

+---+      +-------------+
| b | ---> | value for b |
+---+      +-------------+

Then after the initialization of temp you will have this:

+---+            +-------------+
| a | -----+---> | value for a |
+---+      |     +-------------+
           |
+------+   |
| temp | --/
+------+

+---+      +-------------+
| b | ---> | value for b |
+---+      +-------------+

You have two pointers, pointing to the same location.

Then you do the assignment

*a = *b;

which leads to this situation:

+---+            +-------------+
| a | -----+---> | value for b |
+---+      |     +-------------+
           |
+------+   |
| temp | --/
+------+

+---+      +-------------+
| b | ---> | value for b |
+---+      +-------------+

As can easily be seen, the value for a is lost.

---

PS.

Whenever you have troubles with pointers, and can't really visualize what's happening, I recommend you take a step back, fetch a pencil and paper, and draw it all out.