Misunderstanding something about pointers?

Misunderstanding something about pointers?

• Dejjan

  Because the CNode* is passed by value, meaning it's a copy of the original pointer.  Whatever happens to the copy has no effect on the original.  It's the same as something like this.
  
  {
      int i = 20;
      some_func( i );
  // at this point 'i' is still 20 because the value was passed to the function, not the actual variable.
  }
  
  void some_func( int i ){
      i = 10;
  }
  
  

• DentonChris

  Ah, I did not think of that. So used to the fact that values can be changed so i just never thought about that the pointer itself could not be changed. thank you very much for a quick answer and for teaching me something new!
  
  
  

• Lesley-Ann Vaughan

  This is exactly the same as doing the following:
  
  void f(int i)
  {
     i += 2;
  }
  
  int main()
  {
     int i = 0;
  
     f(i);
  }
  
  The variable 'i' is passed by value and hence any change you make to it within the function f will not be reflected in the calling function.
  
  In C++ pointers are also passed by value: so while you can change the object that the pointer points to you cannot change the pointer itself. If you want to change the pointer then you need to pass it by reference. For example:
  
  void f(X*& pX)
  {
     pX = new X();
  }
  
  int main()
  {
     X* pX = NULL;
  
     f(pX);
  }
  
  

• Pairair

  So how would i go about to avoid this
  
• Misunderstanding something about pointers?