Why sort(rbegin(), rend()) sort the structure on descending order?

Question

for example the code bellow sort the vec on desc order :

std::vector<int> vec = {1, 2, 5, 4, 3};
sort(vec.rbegin(), vec.rend());
for(const auto v : vec)
    std::cout << v << "\n";
output 5 4 3 2 1

On the C++ reference:

Sorts the elements in the range [first,last) into ascending order. The elements are compared using operator< for the first version [...]

Answer 1

The function call indeed sorts the vector in the ascending order starting form the last element up to the first element because there are used reverse iterators

sort(vec.rbegin(), vec.rend());

When you are using reverse iterators then the vector is traversed in the reverse order.

Consider this for loop

for (auto first = vec.rbegin(), last = vec.rend(); first != last; ++first)
{
    std::cout << *first << ' ';
}
std::cout << '\n';

Its output is

3 4 5 2 1

If you want to sort the vector in the ascending order in the direct direction then do not use the reverse iterators. For example

sort(vec.begin(), vec.end());

Or if to use the reverse iterators then instead of the default function object of the type std::less<int> you need to use a function object of the type std::greater<int>

sort( vec.rbegin(), vec.rend(), std::greater<int>() );