macros - Is there a way to get function name inside a C++ function?

Question

I want to implement a function tracer, which would trace how much time a function is taking to execute. I have following class for the same:-

class FuncTracer
{
    public:
        FuncTracer(LPCTSTR strFuncName_in)
        {
            m_strFuncName[0] = _T('');
            if( strFuncName_in ||
                _T('') != strFuncName_in[0])
            {   
                _tcscpy(m_strFuncName,strFuncName_in);

                TCHAR strLog[MAX_PATH];
                _stprintf(strLog,_T("Entering Func:- <%s>"),m_strFuncName);
                LOG(strLog)

                m_dwEnterTime = GetTickCount();
            }
        }

        ~FuncTracer()
        {
            TCHAR strLog[MAX_PATH];
            _stprintf(strLog,_T("Leaving Func:- <%s>, Time inside the func <%d> ms"),m_strFuncName, GetTickCount()-m_dwEnterTime);
            LOG(strLog)
        }

    private:
        TCHAR m_strFuncName[MAX_PATH];
        DWORD m_dwEnterTime;
};

void TestClass::TestFunction()
{
    // I want to avoid writing the function name maually..
    // Is there any macro (__LINE__)or some other way to 
    // get the function name inside a function ??

    FuncTracer(_T("TestClass::TestFunction"));
    /*
     * Rest of the function code.
     */
}

I want to know if there is any way to get the name of the function from inside of a function? Basically I want the users of my class to simply create an object the same. They may not pass the function name.

Answer 1

C99 has __func__, but for C++ this will be compiler specific. On the plus side, some of the compiler-specific versions provide additional type information, which is particularly nice when you're tracing inside a templatized function/class.

• MSVC: __FUNCTION__, __FUNCDNAME__, __FUNCSIG__
• GCC: __func__, __FUNCTION__, __PRETTY_FUNCTION__

Boost library has defined macro BOOST_CURRENT_FUNCTION for most C++ compilers in header boost/current_function.hpp. If the compiler is too old to support this, the result will be "(unknown)".