Compile error with template specialization, default arguments and VS2013 - c++

Question

template<typename T>
void f(const T &v = T());
template<>
void f<std::string>(const std::string &v)
{
std::cout << v;
}
int main(int argc, char* argv[])
{
f<std::string>(); // Error in VS2013, OK in VS2012, gcc-4.7
f<std::string>("Test"); // OK
f<std::string>(std::string()); //OK
return 0;
}
The latest Visual Studio 2013 compiler gives the following compiler error for the case when the default argument must be used:
error C2440: 'default argument' : cannot convert from 'const std::string *' to 'const std::string &'
Reason: cannot convert from 'const std::string *' to 'const std::string'
No constructor could take the source type, or constructor overload resolution was ambiguous
Visual Studio 2012 and gcc-4.7 compile fine.
Update: As it seems to be a VS2013 bug, are there any temporary workarounds that do not require significant code changes until this is fixed by MS? Bug report was submitted on MS connect.

Answer 1

Whenever I see this kind of problems with template functions, I try to switch to template structures (if you need a temporary workaround)...
template<typename T>
struct foo
{
static void f(const T &v = T());
};
template<>
struct foo<std::string>
{
static void f(const std::string &v = std::string())
{
std::cout << v;
}
};
Unfortunately, I can't check this in Visual Studio 2013 because I don't have it, but I hope it should work.
The downside here is that you should explicitly specify your type, it's no longer deducted
foo<std::string>::f()
foo<std::string>::f("Text")
My wild guess here would be adding like a wrapper function:
template<typename T>
void f_wrapper(const T &v = T())
{
foo<T>::f(v);
}