C++ LANGUAGE POINTER TO AN ARRAY

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C++ LANGUAGE POINTER TO AN ARRAY

Pointer to an Array 


It is most likely that you would not understand this chapter 
until you go through the chapter related C++ Pointers. 
So assuming you have bit understanding on pointers in C++, 
let us start: An array name is a constant pointer to the first 
element of the array. Therefore, in the declaration: 

double balance[50]; 

balance is a pointer to &balance[0], which is the address 
of the first element of the array balance. Thus, the following 
program fragment assigns p the address of the first element ofbalance: 

double *p; 

double balance[10]; 
 
p = balance; 

It is legal to use array names as constant pointers, 
and vice versa. Therefore, *(balance + 4) is a legitimate 
way of accessing the data at balance[4]. 

Once you store the address of first element in p, 
you can access array elements using *p, *(p+1), 
*(p+2) and so on. Below is the example to show all 
the concepts discussed above: 

#include <iostream> 

using namespace std; 
  
int main () 
{ 
   // an array with 5 elements. 
   double balance[5] = {1000.0, 2.0, 3.4, 17.0, 50.0}; 
   double *p; 
 
   p = balance; 
  
   // output each array element's value  
   cout << "Array values using pointer " << endl;  
   for ( int i = 0; i < 5; i++ ) 
   { 
       cout << "*(p + " << i << ") : "; 
       cout << *(p + i) << endl; 
   } 
 
   cout << "Array values using balance as address " << endl; 
   for ( int i = 0; i < 5; i++ ) 
   { 
       cout << "*(balance + " << i << ") : "; 
       cout << *(balance + i) << endl; 
   } 
  
   return 0; 
} 
When the above code is compiled and executed, it produces the following result: 
Array values using pointer 
*(p + 0) : 1000 
*(p + 1) : 2 
*(p + 2) : 3.4 
*(p + 3) : 17 
*(p + 4) : 50 
Array values using balance as address 
*(balance + 0) : 1000 
*(balance + 1) : 2 
*(balance + 2) : 3.4 
*(balance + 3) : 17 
*(balance + 4) : 50 
In the above example, p is a pointer to double which means 
it can store address of a variable of double type. Once we 
have address in p, then *p will give us value available at 
the address stored in p, as we have shown in the above example.