why am I getting '1' as last element of the array at index arr[2] after shifting the eleme-CodePudding

why am I getting '1' as last output at index arr[2] after shifting the elements

#include <iostream>
using namespace std;

void shifting(int* arr)
{
    int i, j;
    for (i = 0; i < 3; i  )
    {
        arr[i] = arr[i   1];
    }

    for (i = 0; i < 3; i  )
    {
        cout << arr[i] << endl;
    }
}

int main()
{
    int array[n] = { 5, 2, 3 };
    shifting(array);       //shifting the elements to left side 
    return 0;
}

output: 2 3 1

CodePudding user response：

The way you are shifting is wrong. The proper way to do it is to store the value in a temp element and then shift.

For eg.

temp=arr[0]
for (i = 0; i < 3; i  )
    {
        arr[i] = arr[i   1];
    }

Here add arr[2]=temp

CodePudding user response：

You're indexing an out of bounds array on line

arr[i] = arr[i 1];

when i=2 the right hand side evaluates to arr[3], which is out of bounds.

This generates undefined, buggy behavior. This has already been answered much better by people at this question.

CodePudding user response：

This is called undefined behavior in C , the way you are shifting is indexing the array out of bounds, meaning that you are trying to access an element that doesn't exist in the array.

CodePudding user response：

It depends on what you want the last value to be. If you want the array to output 2, 3, 5:

store the first element(arr[0]) in a temp variable
iterate through the array just like you did but stop at the (n-1)th element i.e., (i < n-1)
then store the temp value in arr[n-1].