Reversing an array (vector) using reverse_iterator - c++

Question

I'm trying to improve my c++ knowledge using Edabit challenges and testing out 'new' functions. Current challenge wants us to write a function to reverse an array.
examples:
reverse([1, 2, 3, 4]) ➞ [4, 3, 2, 1]
reverse([9, 9, 2, 3, 4]) ➞ [4, 3, 2, 9, 9]
reverse([]) ➞ []
I tried to do this using reverse_iterator, but I have some odd result.
std::vector<int> reverse(std::vector<int> arr) {
std::vector<int> out;
for(std::vector<int>::reverse_iterator i = arr.rbegin(); i != arr.rend(); ++i)
{
out.push_back(arr[*i]);
}
return out;
}
The result I get is the following:
FAILED: Expected: equal to [ 4, 3, 2, 1 ]
Actual: [ 0, 4, 3, 2 ]
First it seems like it was just one off, o I added 1 with rbegin() and rend as follows:
for(std::vector<int>::reverse_iterator i = arr.rbegin() + 1; i != arr.rend() + 1; ++i)
And it seems that the first test passed, but following test seems way off:
FAILED: Expected: equal to [ 7, 6, 5 ]
Actual: [ 33, 0, 5 ]
I'm trying to understand how this reverse iterator is working, so please provide a solution and some clear explanation why the first test passed with the addition of 1 and why the second did not pass.

Answer 1

Try:
for(std::vector<int>::reverse_iterator i = arr.rbegin(); i != arr.rend(); ++i)
{
out.push_back(*i);
}
Iterator already gives you a value, not an index into a vector.

Answer 2

The expression in the subscript operator
out.push_back(arr[*i]);
^^^^^^^
does not make sense.
You mean
out.push_back( *i );
But in any case the function can be written better either like this to create a new reversed vector
#include <iostream>
#include <vector>
#include <iterator>
std::vector<int> reverse( const std::vector<int> &v )
{
std::vector<int> out;
out.reserve( v.size() );
for ( auto first = std::rbegin( v ), last = std::rend( v ); first != last; ++first )
{
out.push_back( *first );
}
return out;
}
int main()
{
std::vector<int> v = { 4, 3, 2, 1 };
for ( const auto &item : v ) std::cout << item << ' ';
std::cout << '\n';
auto reversed_v = reverse( v );
for ( const auto &item : reversed_v ) std::cout << item << ' ';
std::cout << '\n';
return 0;
}
The program output is
4 3 2 1
1 2 3 4
Or like this to reverse a vector in place
#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>
void reverse( std::vector<int> &v )
{
if ( not v.empty() )
{
for ( auto first = std::rbegin( v ), last = std::rend( v ); first < --last; ++first )
{
std::iter_swap( first, last );
}
}
}
int main()
{
std::vector<int> v = { 4, 3, 2, 1 };
for ( const auto &item : v ) std::cout << item << ' ';
std::cout << '\n';
reverse( v );
for ( const auto &item : v ) std::cout << item << ' ';
std::cout << '\n';
return 0;
}
The program output is the same as shown above
4 3 2 1
1 2 3 4
Alternatively there are corresponding standard algorithms std::reverse_copy and std::reverse that you can use instead.

Answer 3

Prefer algorithms in the standard library to hand crafted loops as much as possible because:
1) they are more expressive;
2) they are likely to be more efficient.
std::reverse(std::begin(arr), std::end(arr));
Just include the 'algorithm' header to use std::reverse.