# Why array index starts from zero ?

Prerequisite : Pointers in C/C++

Consider int arr[100]. The answer lies in the fact how the compiler interprets arr[i] ( 0<=i<100).

arr[i] is interpreted as *(arr + i). Now, arr is the address of array or address of 0th index element of array. So, address of next element in array is arr + 1 (because elements in array are stored in consecutive memory locations), further address of next location is arr + 2 and so on . Going with above arguments, arr + i means address at i distance away from starting element of array. Therefore, going by this definition, i will be zero for starting element of array because starting element is at 0 distance away from starting element of array. To fit this definition of arr[i], indexing of array starts from 0.

`#include<iostream> ` `using` `namespace` `std; ` ` ` `int` `main() ` `{ ` ` ` `int` `arr[] = {1, 2, 3, 4}; ` ` ` ` ` `// Below two statements mean same thing ` ` ` `cout << *(arr + 1) << ` `" "` `; ` ` ` `cout << arr[1] << ` `" "` `; ` ` ` ` ` `return` `0; ` `} ` |

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**Output:**

2 2

The conclusion is, we need random access in array. To provide random access, compilers use pointer arithmetic to reach i-th element.

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