Question

我正在尝试创建一个奇怪的类，但是它的语法有问题。

#include <memory>
#include <iostream>
#include <string>
#include <typeinfo>
#include <map>

typedef std::shared_ptr<void> SharedVoidPointer;
typedef std::map <std::string, SharedVoidPointer > SharedPtrMap;

template<typename Ts = SharedPtrMap >
class CVar
{
public:
    CVar() : mytype(typeid(Ts)) { //Create ourself as a Shared void pointer containing a Ts
        Ts* var = new Ts();
        ptr = SharedVoidPointer(var);
    }

    CVar(Ts& t) : mytype(typeid(Ts)) { //When we pass a type along to store, we now create ourself as a shared ptr of that type.
        ptr = std::make_shared<Ts>(t);
    }

    template <typename T>
    inline T& operator[] (const std::string& key) {
        if (is<SharedPtrMap>()) { //We are a pointer map. Safe to continue
            SharedPtrMap map = reinterpret_cast<SharedPtrMap>(ptr)[key]; //ptr is a map, so cast
            SharedPtrMap::const_iterator it = map.find(key); //Check for existing key
            if (it != map.end()) { //It already exists.
                return NULL; //TODO: Compare types see if can be reassigned.
            }
            else{
                return reinterpret_cast<T&>(map<T>[key]) ; //Probly not correct
            }
        }
        else{
            return NULL; //Pass error up.
        }
    }

    template <typename T>
    bool is(){
        if (typeid(T) == mytype){
            return true;
        }
        else{
            return false;
        }
    }
private:
    std::shared_ptr<void> ptr;
    const std::type_info& mytype;
};

现在在我的示例程序中：

int main(){
    CVar<> vars;

    if (vars.is<std::string>())
        std::cout << "Should not be string";
    if (vars.is<SharedPtrMap>())
        std::cout << "Should be map.";

    std::string foo = "bar";
    vars["foo"] = foo; //No operator [] matches arguments. Wha?
    vars["foo"]<std::string&>(); Also a no go.
    vars[]<std::string&>("foo"); //Even explicitly typing it does not work.

    foo = "newbar";

    std::cout << vars<std::string>["foo"]; //Should print newbar
    return 0;
}

我对这堂课的期望如下：

vars["something"] = myvariablename;
vars["texture.bg"] = "mytexture.png";
vars["texture.fg"] = "mytexture2.png";

!   Expectations:
vars["something"].is<std::string>()                 //true
vars["something"].is<std::int>()                    //false
vars["somthingelse"].is<std::string>()              //false
vars["somthingelse"].exists()                       //false
vars["texture"].exists()                            //true
vars["texture"].is<SharedPtrMap>                    //true
vars["texture"]                                     //Gives a Cvar<std::map<>>
vars["texture"]["bg"]                               //Gives std::string& ("mytexture.png")
vars["texture.bg"]                                  //Gives std::string& ("mytexture.png")

我的问题似乎与下标运算符的重载有关。它真的不喜欢模板。我假设语法错误。

我也知道我没有“。”分离工作尚未实施。 Kinda挂在这台ATM上。

在 C++ Array Subscript Operator Template

调用操作员的唯一方法是显式编写list.operator []（）。

对此进行了尝试，但即使这样也不起作用。

模板下标运算符重载

0 个答案: