G++ doesn't permit use of protected default constructor in base class when both are templates?

问题

I've created a header for optionally-lazy parameters (also visible in a GitHub repository).

In my original version of the code, I provided a protected default constructor for my base-class template:

template <typename VAL_TYPE>
class LazyType_Base
{
    // ....
    LazyType_Base(void) =default;
// ....

Then, in one of the derived classes:

template <typename VAL_TYPE>
class LazyType_Eager : public LazyType_Base<VAL_TYPE>
{
  public:
    LazyType_Eager(
        VAL_TYPE&& final_val)
      : LazyType_Base<VAL_TYPE>{}
      , val_{final_val}
    {}
// .....

This compiles just fine with Clang++, but in G++ 5.1, I get this error:

In file included from Test_OptionallyLazy.cpp:3:0:
OptionallyLazy.hpp: In instantiation of ‘LazyType_Eager<VAL_TYPE>::LazyType_Eager(VAL_TYPE&&) [with VAL_TYPE = int]’:
Test_OptionallyLazy.cpp:22:14:   required from here
OptionallyLazy.hpp:23:5: error: ‘LazyType_Base<VAL_TYPE>::LazyType_Base() [with VAL_TYPE = int]’ is protected
     LazyType_Base(void) =default;
     ^
OptionallyLazy.hpp:58:23: error: within this context
       , val_{final_val}

What's going on here? The weirdest bit is that the other derived class does something similar that doesn't trigger an error.

Replacing the =default constructor with the explicit default implementation {} resolves the compiler error.

EDIT: Thanks to T.C., here is a real MCVE:

class Meow
{
  protected:
    Meow(void) =default;    
  public:    
    virtual void f() {}
};

class Purr : public Meow
{
  public:
    Purr()
      : Meow{}
    {}

};

来源：https://stackoverflow.com/questions/38213809/g-doesnt-permit-use-of-protected-default-constructor-in-base-class-when-both