New integer array shifted to the right? - c++

Question

I should input the array x[ ] and create the new array y[ ], that is shifted to the right from the original.
I'm havong trouble with the first element, y[0]
for ex. if i input {7, 8, 4, 2, 3}
the code below returns: {-693726240, 7, 8, 4, 2}
where is the error?
Thanks in advance.
#include <iostream>
using namespace std;
int main()
{
int i, j, n, m=0, x[100],y[100];
cin >>n;
for(int i=0; i<n; i++)
{
cout << "enter "<< i+1 <<". element: ";
cin >> x[i];
if (i>0)
y[m++] = x[i-1];
else //THIS IS FOR Y[0] BUT DOESN'T WORK
y[m++] = x[n-1];
}
// ARRAY SHIFTED TO THE RIGHT:
for(int j=0; j<m; j++)
{
cout << y[j]<<" ";
}
return 0;
}

Answer 1

Loop starts with i == 0. You read the i-th element at i-th iteration. i == n - 1 is yet some time in the future when you read x[n - 1] in the 0-th iteration. C++ is not obliged to initialise variables for you to any particular value, so when you read from a not-yet-initialised location, you are likely to get garbage (in this case, -693726240).
To solve this, you have to solve the time paradox. Don't assign anything at iteration 0; wait till the loop is done and all data is in, and only then sneak back and fill in the 0-th element.
Alternately, read the entirety of x first, then when you know all the values, use your current algorithm to shift x into y (i.e. separate the current do-everything loop into a read loop and shift loop). This way, all the x values will be known when you start on y, and you won't have any time-travel mishaps.

Answer 2

I am assuming that you are looking for a circular shift given that you wrote y[m++] = x[n-1]
Amadan's answer highlights a way of solving your problem, here is some code you can follow :
for(int i = 0;i < n-1;i++){
cin>>x[i];`
y[i+1] = x[i];
}
cin>>x[n-1];
y[0] = x[n-1];

Answer 3

got it, great, #Amadan thanks for explanation and quick answer
I also removed "j" and "m" as separate variables for the new array,
think it's not needed actually, I can use the same "i" and "n" that are used for the original array (right? Or should I keep the new variables?)
Here's corrected code that works:
#include <iostream>
using namespace std;
int main()
{
int i, j, n, m=0, x[100],y[100];
cin >>n;
for(int i=0; i<n; i++)
{
cout << "enter "<< i+1 <<". element: ";
cin >> x[i];
if (i>0)
y[i] = x[i-1];
}
y[0] = x[n-1];
// ARRAY SHIFTED TO THE RIGHT:
for(int i=0; i<n; i++)
{
cout << y[i]<<" ";
}
return 0;
}

Answer 4

also I tried the other alternative #Amadan suggested, and used it also for left shifting
- first input all the elements, then assigning another array elements when they're all known
I consider this as the best one, tested and working. Thanks again:
#include <iostream>
using namespace std;
int main()
{
int i, n, x[100],y[100], z[100];
cin >>n;
for(int i=0; i<n; i++)
cin >> x[i];
// ARRAY SHIFTED TO THE RIGHT:
for(int i=0; i<n; i++)
{
if (i > 0) y[i] = x[i-1];
else y[i] = x[n-1];
cout << y[i]<<" ";
}
cout << endl;
// ARRAY SHIFTED TO THE LEFT:
for(int i=0; i<n; i++)
{
if (i < n-1) z[i] = x[i+1];
else z[i] = x[0];
cout << z[i]<<" ";
}
return 0;
}