std::make_shared as a default argument does not compile

Question

In Visual C++ (2008 and 2010), the following code does not compile with the following error:

#include <memory>

void Foo( std::shared_ptr< int >    test = ::std::make_shared< int >( 5 ) )
{
}

class P
{
    void
    Foo( std::shared_ptr< int > test = ::std::make_shared< int >( 5 ) )
    {
    }
};

error C2039: 'make_shared' : is not a member of '`global namespace''

error C3861: 'make_shared': identifier not found

It is complaining about the definition of P::Foo() not ::Foo().

Does anybody know why it is valid for Foo() to have a default argument with std::make_shared but not P::Foo()?

Answer 1

It looks like a bug in the compiler. Here is the minimal code required to reproduce the problem:

namespace ns
{
    template <typename T>
    class test
    {
    };

    template <typename T>
    test<T> func()
    {
        return test<T>();
    }
}

// Works:
void f(ns::test<int> = ns::func<int>()) { }

class test2
{
    // Doesn't work:
    void g(ns::test<int> = ns::func<int>()) 
    { 
    }
};

Visual C++ 2008 and 2010 both report:

error C2783: 'ns::test<T> ns::func(void)' : could not deduce template argument for 'T'

Comeau has no issues with this code.