C++, vector: getting incorrect output containing 0's - c++

Question

Why is the vector 'r' giving the output as follows? Instead, it should not have zero in the list.
Can someone help?
output: 0 0 0 0 0 5 1 2 3 4
vector <int> leftRotation(vector <int> a, int d) {
vector<int> r(a.size());
// int j=0;
for(int i=d; i<a.size(); i++)
r.push_back(a[i]);
for(int i=0; i<d; i++)
r.push_back(a[i]);
return r;
}

Answer 1

You are initializing r with a.size() copies of 0 at the start of your code:
vector<int> r(a.size());
At this point, the vector r contains values 0 0 0 0 0 (as many zeroes as there are elements in a).
Then, you push values onto the back of this vector of zeroes. At each step of your loops:
0 0 0 0 0 5
0 0 0 0 0 5 1
0 0 0 0 0 5 1 2
0 0 0 0 0 5 1 2 3
0 0 0 0 0 5 1 2 3 4
Instead, initialize with an empty vector:
vector<int> r;

Answer 2

The local variable of vector<int> r(a.size()) uses a vector's constructor overload that accepts (size_type count) argument and is initialized (or rather zeroed out) to contain a.size() number of 0s. For clarification check out the std::vector constructor overload no. 3:
The overload (3) zeroes out elements of non-class types such as int
Use a default constructor to create an empty container instead:
std::vector<int> r;
or:
std::vector<int> r{};

Answer 3

vector<int> r(a.size());
With this line, you create a vector of initial size a.size(); the initial elements in the vector are set to 0.
Probably your a vector has size 5, so this is where the five zeros in your output come from:
0 0 0 0 0
Then, you invoke the push_back() method on r, and this appends other elements at the end of the r vector.
To get a better help, please clarify your goal.