std::bind lose reference when delivered as rvalue reference

Question

I have the following code:

#include 
#include 

template 
auto callback(T&& func) ->decltype(fu         


        

Answer 1

std::bind() is designed for value semantics (as R. Martinho Fernandes nicely explains in his answer), and does create copies internally. What you need/want is std::ref:

callback(std::bind(test, std::ref(t)));
//                       ^^^^^^^^^^^

std::ref returns an std::reference_wrapper<> object that wraps a reference to your original argument. This way, the reference_wrapper object around t gets copied, and not t itself.

This allows you to choose between value semantics (assumed by default) and reference semantics (which requires your explicit intervention).

Here is a live example.