Consider this code:
#include
#include
int xx = 7;
template
void f1(T arg)
{
arg += xx;
}
template
A small modification to f2
provides the clue:
template
void f2(T arg)
{
arg.get() = xx;
}
This now does what you expect.
This has happened because std::ref
returns a std::reference_wrapper<>
object. The assignment operator of which rebinds the wrapper.
(see http://en.cppreference.com/w/cpp/utility/functional/reference_wrapper/operator%3D)
It does not make an assignment to the wrapped reference.
In the f1
case, all is working as you expected because a std::reference_wrapper
provides a conversion operator to T&
, which will bind to the implicit right hand side of int
s implicit operator+
.