An Uncommon representation of array elements
March 4, 2010
Consider the below program.
int main( )
{
int arr[2] = {0,1};
printf("First Element = %d\n",arr[0]);
getchar();
return 0;
}
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.
