强制模板化构造函数运行？

Question

我知道有一个initailzer技巧可以强制构建一个全局对象，无论它在何处使用。这用于std :: cout我相信。

#ifndef GUARD_H
#define GUARD_H

class Magical
{
   // default constructor and such...
};

class Init
{
public:
    Init();
};

extern Magical& magic;

namespace
{
    Init __magical_initializer; // works as this object is constructed in every source file it is included in
}

#endif

的src：

#include "magical.h"

#include <new>

static int count; // believe there is a spec somewhere which states global integers are initialized with zero

static alignas(Magical) char buffer[sizeof(Magical)];

Magical& magic = *reinterpret_cast<Magical*>(buffer);

Init::Init()
{
    if(!count++)
    {
        new(buffer) Magical;
    }
}

我想知道是否有与此相当的模板，因此我的代码看起来像这样：

template<typename T>
class Base
{

    static Magical<T> __private; // need this constructor to be called.

};

// usage:

class SomeClass : public Base<SomeClass>
{
};

Answer 1

无法解决此问题，因为源文件中不存在模板。