C++ Pointers-Concept Explained

Replies

• 

  Deepika Bansal

  Lets consider the following as well:
  &*p is same as &(*p) which means address of the value stored.
  *&*p is same as *(&(*p)) which means value of variable through pointer p.
  *&p is same as *(&p) which means value of pointer.
  
  This means we always resolve such expressions from the inwards and moving outwards.
• 

  Deepika Bansal

  Pointer Expression
  

  int x=5,y=5;
  int *p, *q;
  int a,b,c,d;
  p=&x;               //p now contains the memory address of x ie p refers to x
  q=&y;              //q now contains the memory address of y ie q refers to y
  a=(*p)+(*q);   //x+y
  b=(*p)-(*q);   //x-y
  c=(*p)*(*q);   //x*y
   d=(*p)/(*q);   //x/y
  *p=*p+10;     // x=x+10
  

  Here we are accessing the value of x and y using the pointers p and q as *p and *q respectively.
• 

  Deepika Bansal

  Scale Factor
  We can also have something like p++ or p--.
  

  int *p;
  int x=10;
  p=&x;
  p++;
  

  Here when we are incrementing or decrementing the pointer variable p is according to the data type of the variable it is pointing to i.e. x in our case.
  In our example we are considering x to be of type int. Int type requires 2 bytes of memory location to store one integer value. So when we're having p++, we are actually moving two bytes so that p may now point to next integer value (which may be a garbage value as well).
  
  Had p be of type char *, p would have been incremeted by 1 byte to point to next char.
  Similarly for float *, p would have been incremented by 4 bytes to point to next float and so on.
  
  So the key formula is:
  ptr++ is same as ptr=ptr+1 which in turn is equivalent to ptr=ptr + (1*size of element).
  
  PS: Division and Multiplication is not applicable this way.