C++ Pointer to Array
You can skip this chapter and come back to it after understanding the concept of C++ pointers.
If you are familiar with the concept of C++ pointers, you can start learning this chapter. The array name is a constant pointer to the first element of the array. Therefore, in the following declaration:
double tutorialproAarray[50];
tutorialproAarray is a pointer to &tutorialproAarray[0], which is the address of the first element of the array tutorialproAarray. Thus, the following snippet assigns the address of the first element of tutorialproAarray to p:
double *p;
double tutorialproAarray[10];
p = tutorialproAarray;
Using the array name as a constant pointer is legal, and vice versa. Therefore, *(tutorialproAarray + 4) is a legal way to access the data in tutorialproAarray[4].
Once you store the address of the first element in p, you can access the array elements using *p, *(p+1), *(p+2), etc. The following example demonstrates these concepts:
Example
#include <iostream>
using namespace std;
int main ()
{
// Array of 5 double elements
double tutorialproAarray[5] = {1000.0, 2.0, 3.4, 17.0, 50.0};
double *p;
p = tutorialproAarray;
// Print the values of each element in the array
cout << "Array values using pointer " << endl;
for ( int i = 0; i < 5; i++ )
{
cout << "*(p + " << i << ") : ";
cout << *(p + i) << endl;
}
cout << "Array values using tutorialproAarray as address " << endl;
for ( int i = 0; i < 5; i++ )
{
cout << "*(tutorialproAarray + " << i << ") : ";
cout << *(tutorialproAarray + 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 tutorialproAarray as address
*(tutorialproAarray + 0) : 1000
*(tutorialproAarray + 1) : 2
*(tutorialproAarray + 2) : 3.4
*(tutorialproAarray + 3) : 17
*(tutorialproAarray + 4) : 50
In the above example, p is a pointer to a double, which means it can store the address of a double variable. Once we have the address in p, *p will give the value stored at the corresponding address in p, as demonstrated in the example above.