Consider the below program.
C
int main( )
{
int arr[2] = {0,1};
printf ( "First Element = %d\n" ,arr[0]);
getchar ();
return 0;
}
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C++
#include <iostream>
using namespace std;
int main() {
int arr[2] = {0, 1};
cout << "First Element = " << 0[arr] << endl;
return 0;
}
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Pretty Simple program.. huh… Output will be 0.
Now if you replace arr[0] with 0[arr], the output would be same. Because compiler converts the array operation in pointers before accessing the array elements. e.g. arr[0] would be *(arr + 0) and therefore 0[arr] would be *(0 + arr) and you know that both *(arr + 0) and *(0 + arr) are same.
Please write comments if you find anything incorrect in the above article.