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.

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#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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