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An Uncommon representation of array elements
  • Difficulty Level : Easy
  • Last Updated : 09 Nov, 2020

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.

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