In much the same way that a pointer can be const:

    int *const p;

I propose that you can apply the 'explicit' modifier to an R-value reference:

    int &&explicit x;

Consider the following code snippet:

#include <iostream>  // cout, endl
#include <utility>   // move

using namespace std;

void Func(int&) { cout << "L-value reference" << endl; }

void Func(int const &) { cout << "L-value reference to const" << endl; }

void Func(int&&) { cout << "R-value reference" << endl; }

void Func(int const &&) { cout << "R-value reference to const" << endl; }

int main(int argc, char **argv)
{
    int val = 7;

    int &&explicit x = move(val);

    Func(x);   // This prints 'R-value reference'
}

So the effect of the 'explicit' modifier on an R-value reference is
that it gets treated as an R-value reference from that point onward
(rather than being treated as an L-value reference).

You could use it in function parameters as well:

void SetVector( vector<int> &&explicit arg )
{
    static vector<int> vec;
    vec = arg;   // no need for 'move' here
}

Furthermore, when used with a template, it could eradicate the need
for "std::forward" as follows:

template<typename T>
void SetVector( T &&explicit arg )
{
    static T obj;
    vec = arg;   // no need for 'forward' here
}