获取的模板类型的类型名，没有类定义 [英] Obtaining the type-name of a template type, without class definition

问题描述

我试图写一个模板，包装与运作的 smart_ptr 键入并需要抛出一个异常，在某些情况下。对于这种情况，我想包括类包装类型的名称。由于我与智能指针工作只能向前声明将提供类型。

I am trying to write a template wrapper that works with the smart_ptr type and need to throw an exception in some cases. For this case I'd like to include the name of the type the class is wrapping. As I am working with smart pointers only forward declarations will be available for the type.

所以，重要的问题是，我怎么能得到一个字符串模板参数，而无需其定义可用？ （我并不需要一个干净的名字，任何类似的名称将被罚款）

So the essential question is how can I get a string for of a template parameter without having its definition available? (I don't need a clean name, anything resembling the name will be fine)

我在尝试使用的typeid 失败，因为它需要的类定义（至少在GCC）。

My attempt at using typeid fails since it requires the class definition (at least in GCC).

在code基本上我需要努力的是低于（这给出了GCC错误）

The code I essentially need to work is below (which gives an error in GCC)

#include <boost/shared_ptr.hpp>

using boost::shared_ptr;
class SomeClass;

void func( shared_ptr<SomeClass> obj );

template<class T>
class Wrap
{
    shared_ptr<T> ptr;
public:
    shared_ptr<T> get()
    {
        if( !ptr )
            throw std::string(typeid(T).name());
        return ptr;
    }
};

extern Wrap<SomeClass> wrapSomeClass;

int main()
{
    func( wrapSomeClass.get() );
}

的（这种设置使我生病态previous <一个href=\"http://stackoverflow.com/questions/6523292/forwarding-a-shared-ptr-without-class-declaration\">question - 错误信息是怎么样的混乱）的

(This setup led to my ill-stated previous question -- the error message is kind of confusing)

推荐答案

您可以写一个code发生器创建一个模板，如 TYPE_STRING 专门用于所有需要的类型，并使用该模板来获得一个字符串。我想不出，这并不需要完整的定义，任何其他方式。

You could write a code generator that creates a template such as type_string specialized for all needed types and use that template to get a string. I can't think of any other way that doesn't need the full definition.

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