Shifting an array to the right - c++

Question

I'm generating an array of random integers and trying to shift the values one to the right and replace the first element with the former last element.
The output is not ordered, and the final element is a randomly generated integer.
#include <iostream>
#include <cstdlib>
#include <iomanip>
using namespace std;
void shift(int values[], int size) {
int temp;
for (int i = 0; i < size; i++) {
temp = values[size - 1];
values[i] = values[i + 1];
values[0] = temp;
cout << values[i] << setw(4);
}
cout << endl;
}
int main()
{
cout << "Random 10 index array" << endl;
const int CAP = 10;
int numbers[CAP];
srand(time(0));
int i;
for (i = 0; i < CAP; i++) {
int rng = rand() % 100 + 1;
numbers[i] = rng;
cout << numbers[i] << setw(4);
}
cout << "shifting all elements to the right: " << endl;
shift(numbers, CAP);
cout << endl;
system("pause");
return 0;
}
I tried using i < size - 1, but I got 9 out of the 10 numbers I needed.

Answer 1

Have you tried
If you want a circular shift of the elements:
std::rotate(&arr[0], &arr1, &arr[10]); ... will do the trick. You'll
need to #include the algorithm header.
Optimal way to perform a shift operation on an array
Edit: As pointed out, std::rotate rotates left if used directly. Here is an example doing rotate right on a vector with some change:
#include <vector>
#include <iostream>
#include <algorithm>
int main()
{
std::vector<int> v{2, 4, 2, 0, 5, 10, 7, 3, 7, 1};
// simple rotation to the right
std::rotate(v.rbegin(), v.rbegin() + 1, v.rend());
std::cout << "simple rotate right : ";
for (int n: v)
std::cout << n << ' ';
std::cout << '\n';
}
Output:
simple rotate right : 1 2 4 2 0 5 10 7 3 7

Answer 2

Here's the offending code:
temp = values[size - 1];
This statement does not use the loop variable. Why is it sitting in the loop? This assignment will keep happening size-1 times.
values[i] = values[i + 1];
Your loop invariant is i <size, yet you try to access i+1. That's just asking for trouble :). This is why you don't get garbage values when you use i < size-1.
values[0] = temp;
Again, this doesn't use the loop variable. It doesn't belong in the loop. You just keep setting values[0]over and over again.
Here's a solution that works, using two temp variables:
void shift(int values[], int size) {
7 int temp = values[size-1], temp1;
8 for (int i = 0; i < size; i++) {
9 temp1 = values[i];
10 values[i] = temp;
11 temp = temp1;
12 cout << values[i] << setw(4);
13 }
14 cout << endl;
15 }

Answer 3

Another soution would be to copy the end of the array to the beginning of a temporary array and copy the beginning to the end of the temporary. Here is an exapmle with std::vector.
#include <algorithm>
#include <vector>
vector<int> shiftRight(vector<int> &A, int K)
{
if (A.size() <= 1) // No need to rotate if the array is smaller than 2 elements
return A;
K %= A.size(); // ensure K is less than size.
vector<int> S(A.size());
std::copy(A.cend() - K, A.cend() , S.begin());
std::copy(A.cbegin(), A.cend() - K, S.begin() + K);
return S;
}
It can be also implemented with memcpy which will be faster for big arrays.
#include <cstring>
void shiftRight(int *A, int size, int K)
{
if (size <= 1) // No need to rotate if the array is smaller than 2 elements
return;
K %= size; // ensure K is less than size.
int *temp = new int[size];
memcpy(temp, A + (size - K), K*sizeof(int));
memcpy(temp + K, A, (size - K)*sizeof(int));
memcpy(A, temp, size*sizeof(int)); // copy back to the original array
delete [] temp;
}