C++ LANGUAGE POINTERS VS ARRAYS

COMPUTER LANGUAGES HTML,C,C++.JAVA,.NET AND MULTIMEDIA basics and programs click home button

C++ LANGUAGE POINTERS VS ARRAYS

Pointers vs Arrays 


Pointers and arrays are strongly related. In fact, pointers 
and arrays are interchangeable in many cases. For example, 
a pointer that points to the beginning of an array can access 
that array by using either pointer arithmetic or array-style 
indexing. Consider the following program: 

#include <iostream> 
  
using namespace std; 
const int MAX = 3; 
  
int main () 
{ 
   int  var[MAX] = {10, 100, 200}; 
   int  *ptr; 
  
   // let us have array address in pointer. 
   ptr = var; 
   for (int i = 0; i < MAX; i++) 
   { 
      cout << "Address of var[" << i << "] = "; 
      cout << ptr << endl; 
  
      cout << "Value of var[" << i << "] = "; 
      cout << *ptr << endl; 
  
      // point to the next location 
      ptr++; 
   } 
   return 0; 
} 
When the above code is compiled and executed, it produces result something as follows: 
Address of var[0] = 0xbfa088b0 
Value of var[0] = 10 
Address of var[1] = 0xbfa088b4 
Value of var[1] = 100 
Address of var[2] = 0xbfa088b8 
Value of var[2] = 200 
However, pointers and arrays are not completely interchangeable. For example, consider the following program: 
#include <iostream> 
  
using namespace std; 
const int MAX = 3; 
  
int main () 
{ 
   int  var[MAX] = {10, 100, 200}; 
  
   for (int i = 0; i < MAX; i++) 
   { 
      *var = i;    // This is a correct syntax 
      var++;       // This is incorrect. 
   } 
   return 0; 
} 
It is perfectly acceptable to apply the pointer operator * to var but it is illegal to modify var value. The reason for this is that var is a constant that points to the beginning of an array and can not be used as l-value. 
Because an array name generates a pointer constant, it can still be used in pointer-style expressions, as long as it is not modified. For example, the following is a valid statement that assigns var[2] the value 500: 
*(var + 2) = 500; 
Above statement is valid and will compile successfully because var is not changed.