奇怪的重复模板模式（CRTP）和派生的构造函数参数

Question

I'm using curiously recurring template pattern for creating shared pointer in the following way (below). On Derived::create(...) Visual Studio IntelliSense shows than available arguments are (Args &&...args). How to pass Derived class constructor argument list to Base so that IntelliSense would show me that available arguments are (const std::string &str, int i)?

#include <memory>
#include <string>

template<typename T>
class Base
{
public:
    template<typename... Args >
    static std::shared_ptr<T> create(Args&&... args)
    {
        return std::make_shared<T>(std::forward<Args>(args)...);
    }
};

class Derived : public Base<Derived>
{
public:
    Derived(const std::string &str, int i) {}
};

int main()
{
    auto derived = Derived::create("text", 123);
}

Answer 1

"How to pass Derived class constructor argument list to Base so that IntelliSense would show me that available arguments are (const std::string &str, int i)?"

#include <string>
#include <memory>

template<typename T>
class Base {
public:
    template<typename... Args >
    static std::shared_ptr<T> create(Args&&... args) {
        return std::make_shared<T>(std::forward<Args>(args)...);
    }
};

class Derived : public Base<Derived> {
public:
    Derived(const std::string &str, int i) {}
};

int main() {
    auto derived = Derived::create("text", 123);
}

Well, your code just compiles fine .

Intellisense features of any IDE are always just as good as the c++ parser used for them. That's completly dependent on the IDE actually used, and you shouldn't orient your designs about your IDE's capabilities, but what compiles and works well.