Accessing outer scope name in a lambda, is g++ or Visual or neither, right?

Question

Code:

auto main() -> int
{
    int b = 42;
    auto lambasta = [&]()
    {
        using B_type = decltype( b );
        return B_type{};
    };
    (void) lambasta;
}

Compiles with no diagnostic with MinGW g++ 6.3.0 -std=c++14 -Wall -pedantic-errors. Fails to compile with Visual C++ 2015 update 3,

foo.cpp(6): error C2065: 'b': undeclared identifier

Answer 1

Possible workaround:

template<typename T>
struct wrapper
{
    using wrapped_t = T;
};

auto main() -> int
{
    int b = 42;
    auto lambasta = [&, a = wrapper<decltype(b)>()]()
    {
        using B_type = typename decltype( a ) ::wrapped_t;
        return B_type{};
    };
    (void) lambasta;
}

Works on GCC 6.3 and MSVC 2015 up 3