C++ 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 of balance −

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>usingnamespace std;intmain(){// 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;}return0;}

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.