C++ 17中没有自动参数的模板化lambda

Question

我有一个Derived继承自类的类Base<ResourceType>:

template <class ResourceType>
class Base {
protected:
  ResourceType* resource;
public:
  void set_resource(ResourceType* resource) {
    this->resource = resource;
  }
};

template <class ResourceType>
class Derived : public Base<ResourceType> {
public:
  using Base<ResourceType>::resource;
  void print () {
    std::cout << *resource << std::endl;
  }
};

我想创建一个创建类型对象的工厂Derived.我当然可以用功能做到这一点:

template <typename ResourceType>
auto derived_factory () { 
  return new Derived<ResourceType>(); 
}

auto derived = *(derived_factory<int>());

但是,我无法为工厂编写lambda函数.如果我使用auto关键字接受模板参数,我可以编写模板化的lambda函数,但在这里我只想使用模板来确定返回类型.以下失败:

auto derived_factory = []<typename ResourceType>() {
  return new Derived<ResourceType>();
};

auto derived = *(derived_factory<int>());

有错误:

inherit_unknown_type.cpp: In function ‘int main()’:
inherit_unknown_type.cpp:27:36: error: expected primary-expression before ‘int’
   auto derived = *(derived_factory<int>());
                                    ^~~
inherit_unknown_type.cpp:27:36: error: expected ‘)’ before ‘int’

我只是错误地调用了lambda吗？或者我必须等待C++20？

Answer 1

lambda表达式中的模板参数列表是C++ 20的一个特性.

^{(事实上,我的海湾合作委员会说,在诊断:error: lambda templates are only available with -std=c++2a or -std=gnu++2a [-Wpedantic])}

但是你不必等待C++ 20,GCC 8 已经支持了-std=c++2aflag.

而且你必须改变调用语法:而不是derived_factory<int>(),你需要derived_factory.operator()<int>().

作为替代方案(如果您不想要免费功能),我建议使用标签调度的变体:

auto derived_factory = [](auto tag) {
    return new Derived<typename tag::type>();
};

template <typename T> struct tag_type {using type = T;};

// Usage:
derived_factory(tag_type<int>{})

此外,即使你以某种方式编译它,这一行:

auto derived = *(derived_factory<int>());

无论如何都会导致内存泄漏.为避免这种情况,您应将结果存储为指针或引用.或者甚至更好,使用智能指针.

Answer 2

等待C++ 20,您可以从模板类返回lambda

template <typename ResourceType>
auto make_derived_factory ()
 { return []{ return new Derived<ResourceType>{}; }; }

auto derived = make_derived_factory<int>();

int main ()
 {
   auto df { derived() };
 }