无法拆分头文件和源文件中具有静态成员函数的类

我正在使用一个简单的日志类来调试我的代码。代码已从下载http://www.drdobbs.com/cpp/logging-in-c/201804215?pgno=1并且是完全在头文件中实现的日志类。

由于缺少对象文件而导致的清晰性和构建困难，我想将实现与声明分开。也就是说，将代码实现移动到一个cpp文件中。

一旦我将实现细节移动到.cpp文件中，我最终会得到许多针对所有静态成员函数的"未定义引用"链接器错误。错误源自我的代码的其余部分。从任何地方调用日志记录函数。

该问题是否与.cpp文件中静态成员函数的内部链接有关？如果是，我有哪些选项可以实现文件拆分？一种选择是删除类封装并将其重新实现为非OO代码。我也宁愿避免，还有别的吗？

main.cpp

int main(int argc, char* argv[])
{
    FILELog::ReportingLevel() = FILELog::FromString(argv[1] ? argv[1] : "DEBUG1");
        const int count = 3;
    FILE_LOG(logDEBUG) << "A loop with " << count << " iterations";
    for (int i = 0; i != count; ++i)
    {
            FILE_LOG(logDEBUG1) << "the counter i = " << i;
    }
    return 0;
}

log.hpp

#ifndef __LOG_H__
#define __LOG_H__
#include <sstream>
#include <string>
#include <stdio.h>

#ifndef FILELOG_MAX_LEVEL
#define FILELOG_MAX_LEVEL logDEBUG4
#endif
#define FILE_LOG(level) 
    if (level > FILELOG_MAX_LEVEL) ;
    else if (level > FILELog::ReportingLevel() || !Output2FILE::Stream()) ; 
    else FILELog().Get(level)

enum TLogLevel 
{
    logERROR, 
    logWARNING, 
    logINFO, 
    logDEBUG, 
    logDEBUG1, 
    logDEBUG2, 
    logDEBUG3, 
    logDEBUG4
};
class Output2FILE
{
public:
    static FILE*& Stream();
    static void Output(const std::string& msg);
};

template <typename T>
class Log
{
public:
    Log();
    virtual ~Log();
    std::ostringstream& Get(TLogLevel level = logINFO);
public:
    static TLogLevel& ReportingLevel();
    static std::string ToString(TLogLevel level);
    static TLogLevel FromString(const std::string& level);
protected:
    std::ostringstream os;
private:
    Log(const Log&);
    Log& operator =(const Log&);
};
//class FILELog : public Log<Output2FILE> {};
typedef Log<Output2FILE> FILELog;

#endif //__LOG_H__

log.cpp

#include <sys/time.h>
#include "log.hpp"
inline std::string NowTime()
{
    char buffer[11];
    time_t t;
    time(&t);
    //tm r = {0};
    tm r = {0,0,0,0,0,0,0,0,0,0,0};
    strftime(buffer, sizeof(buffer), "%X", localtime_r(&t, &r));
    struct timeval tv;
    gettimeofday(&tv, 0);
    char result[100] = {0};
    sprintf(result, "%s.%03ld", buffer, (long)tv.tv_usec / 1000); 
    return result;
}
//----------------------------------------------------------------------
template <typename T>
Log<T>::Log()
{
}
template <typename T>
std::ostringstream& Log<T>::Get(TLogLevel level)
{
    os << "- " << NowTime();
    os << " " << ToString(level) << ": ";
    os << std::string(level > logDEBUG ? level - logDEBUG : 0, 't');
    return os;
}
template <typename T>
Log<T>::~Log()
{
    os << std::endl;
    T::Output(os.str());
}
template <typename T>
TLogLevel& Log<T>::ReportingLevel()
{
    static TLogLevel reportingLevel = logDEBUG4;
    return reportingLevel;
}
template <typename T>
std::string Log<T>::ToString(TLogLevel level)
{
    static const char* const buffer[] = {"ERROR", "WARNING", "INFO", "DEBUG", "DEBUG1", "DEBUG2", "DEBUG3", "DEBUG4"};
    return buffer[level];
}
template <typename T>
TLogLevel Log<T>::FromString(const std::string& level)
{
    if (level == "DEBUG4")
        return logDEBUG4;
    if (level == "DEBUG3")
        return logDEBUG3;
    if (level == "DEBUG2")
        return logDEBUG2;
    if (level == "DEBUG1")
        return logDEBUG1;
    if (level == "DEBUG")
        return logDEBUG;
    if (level == "INFO")
        return logINFO;
    if (level == "WARNING")
        return logWARNING;
    if (level == "ERROR")
        return logERROR;
    Log<T>().Get(logWARNING) << "Unknown logging level '" << level << "'. Using INFO level as default.";
    return logINFO;
}
//----------------------------------------------------------------------
inline FILE*& Output2FILE::Stream()
{
    static FILE* pStream = stderr;
    return pStream;
}
inline void Output2FILE::Output(const std::string& msg)
{   
    FILE* pStream = Stream();
    if (!pStream)
        return;
    fprintf(pStream, "%s", msg.c_str());
    fflush(pStream);
}