Thread: Question about this Array

I was just playing around with some code when I ran into this.
The output for this code was 7, 1, 2686728, 9, 7, 5, 3, 1

(if you run this your third number may be different)

In the code I get an address when cout is executed the third time note (anArray - 1). "I thought it would print (anArray) one element below the last one but I was wrong" Is this the address the actual address of element [3] or the address one byte below it or what?
How can I prove what it is??

Code:

#include <iostream>
#include <string>
#include <cmath>
using namespace std;
int main ()
{
int anArray[5] = { 9, 7, 5, 3, 1 };
cout << *(anArray+1) << endl;
cout << * (anArray +4) <<endl;
cout << *(anArray -1) <<endl;
cout << anArray [0]<<endl;
cout << anArray [1] <<endl;
cout << anArray [2] <<endl;
cout << anArray [3] <<endl;
cout << anArray [4] <<endl;
return 0;
}

How can I prove what it is??

You can't. When you try to access an array out of bounds you are in undefined territory.

In the code I get an address when cout is executed the third time note

No, you get an undefined value.

Is this the address the actual address of element [3] or the address one byte below it or what?

No it is not an address, it is whatever value resides at the address of the array - 1.



Jim

Thank You all for the input. It is understandable that anArray [-1] is not part of the array. I should have caught it myself!