转发参数包到构造函数()在g ++ 6.2.1中失败

Question

如何在g ++ - 6.2.1中克服/解决此错误

以下代码适用于g ++ - 7.3.0,但升级编译器对我来说不是一个选项.所以我正在寻找一些SFINAE魔法......尝试很少但到目前为止失败了......

class Base {
public:
  Base(std::string str) : s(std::make_shared<std::string>(str)) {}
  Base(Base &&base) noexcept { s = std::move(base.s); }
  Base &operator=(Base &&i_base_actor) noexcept {
      s = std::move(i_base_actor.s);
      return *this;
  }
  virtual ~Base() = default;

private:
  std::shared_ptr<std::string> s;
};

// Derived
class Derived : public Base {
public:
   Derived() :Base("Derived") {}
   ~Derived() = default;
};

// Derived1
class Derived1 : public Base {
public:
   Derived1(int a) :Base("Derived1") {}
   ~Derived1() = default;
};

包装功能:

template<typename T, typename... Args>
T construct(Args&&... args) {
   return T(std::forward<Args>(args)...);
}

主要:

int main() {
  construct<Derived>();
  construct<Derived1>(100);
}

g ++中的错误

optional_params.derived.cc: In instantiation of ‘T construct(Args&& ...) [with T = Derived; Args = {}]’:
optional_params.derived.cc:42:22:   required from here
optional_params.derived.cc:37:19: error: use of deleted function ‘Derived::Derived(const Derived&)’
   return T(args...);
                   ^
optional_params.derived.cc:21:7: note: ‘Derived::Derived(const Derived&)’ is implicitly deleted because the default definition would be ill-formed:
 class Derived : public Base {
       ^~~~~~~
optional_params.derived.cc:21:7: error: use of deleted function ‘Base::Base(const Base&)’
optional_params.derived.cc:4:7: note: ‘Base::Base(const Base&)’ is implicitly declared as deleted because ‘Base’ declares a move constructor or move assignment operator
 class Base {
       ^~~~

Answer 1

您的代码依赖于以下行中的保证copy elision C++ 17:

template<typename T, typename... Args>
T construct(Args&&... args) {
   return T(std::forward<Args>(args)...);  // <----- copy elison
}

基本上,它表示从C++ 17开始,编译器T在这种情况下不能复制,并且需要在调用者中直接构造它.在C++ 14及更早版本中,编译器必须确保可以访问移动(或复制)构造函数,即使在优化复制构造函数的情况下也是如此.显然,即使有了-std=c++17标志,gcc-6.2.1也不支持C++ 17的这个方面.

最简单的方法是向派生类添加一个移动构造函数:

Derived(Derived &&) noexcept = default;

这样,C++ 14编译器发现,即使在未执行复制省略的假设情况下,也有一种方法可以返回值.请注意,任何合理的C++ 14编译器都将执行复制省略,但它仍将确保可以访问复制或移动构造函数.从C++ 17开始,没有执行这样的测试,因为编译器必须在这种情况下忽略复制/移动.

正如评论部分所述,另一种可能性是:

template<typename T, typename... Args>
T construct(Args&&... args) {
   return {std::forward<Args>(args)...};
}

它也将直接在调用者中构造它,但前提T是构造函数不是显式的.

或者,另一条评论建议避免使用显式析构函数.显式析构函数禁止自动生成默认的move-constructors:

class Derived : public Base {
public:
    Derived() :Base("Derived") {}
   //~Derived() = default; <-- not really needed.
};

但是,由于这只是一个可重复性最小的示例,因此在完整代码中可能需要显式析构函数.在这种情况下,避免析构函数不是一种选择.