Possible to create std::map as rvalue?

Question

I'm not sure if this is an error with my C++ syntax, or if this is something that cannot be accomplished at all.

I want to define a class that takes a std::map as a constructor argument. I then want to create an instance of that class by passing a "temporary" (appropriate to call this "rvalue"?) std::map. I.e. I do not want to create an lvalue std::map and then pass that to the constructor.

Can this be accomplished? I have tried the following (commented lines show failed attempts)

#include <map>
#include <string>
#include <iostream>

class Test
{
    public:
        Test(std::map<int,char>& rMap)
        {
            std::map<int,char>::iterator iter;

            for (iter = rMap.begin(); iter != rMap.end(); ++iter)
            {
                mMap[iter->first] = mMap[iter->second];
            }
        }
        virtual ~Test(){}

    protected:
        std::map<int, char> mMap;
};

int main()
{
    std::cout << "Hello world!" << std::endl;

    //Test test({1,'a'});                   // Compile error.
    //Test test(std::map<int,char>(1,'a')); // Also compile error.
    //Test test(std::map<int,char>{1,'a'}); // Yet again compile error.

    return 0;
}

This is my compiler:

g++ (GCC) 4.4.7 20120313 (Red Hat 4.4.7-11)

Compile errors can be posted upon request, but I'm not sure if they would be useful if my problem is syntactic.

Thank you.

Answer 1

Do

Test(std::map<int, char> rMap) : mMap(std::move(rMap)) {}

or

Test(std::map<int, char>&& rMap) : mMap(std::move(rMap)) {}

or

Test(const std::map<int, char>& rMap) : mMap(rMap) {}

Temporary cannot bind to non const l-value reference.

And use it as

Test test({{1,'a'}});
Test test2({{1,'a'}, {2, 'b'}});

Answer 2

Yes but your constructor takes an lvalue reference. It must instead be a reference-to-const, or an rvalue reference.

Just like with any other type.