customize cout | 易学教程

问题

How can I derive a class from cout so that, for example, writing to it

new_cout << "message";

would be equivalent to

cout << __FUNCTION__ << "message" << "end of message" << endl;

回答1:

class Log
{
public:
    Log(const std::string &funcName)
    {
        std::cout << funcName << ": ";
    }

    template <class T>
    Log &operator<<(const T &v)
    {
        std::cout << v;
        return *this;
    }

    ~Log()
    {
        std::cout << " [end of message]" << std::endl;
    }
};

#define MAGIC_LOG Log(__FUNCTION__)

Hence:

MAGIC_LOG << "here's a message";
MAGIC_LOG << "here's one with a number: " << 5;

回答2:

#define debug_print(message) (std::cout << __FUNCTION__ << (message) << std::endl)

This has the advantage that you can disable all debug messages at once when you're done

#define debug_print(message) ()

回答3:

Further from Mykola's response, I have the following implementation in my code. The usage is

         LOG_DEBUG("print 3 " << 3);

prints

         DEBUG (f.cpp, 101): print 3 3

You can modify it to use FUNCTION along/in place of LINE and FILE

/// Implements a simple logging facility. 
class Logger
{
        std::ostringstream os_;
        static Logger* instance_;
        Logger();
public:
        static Logger* getLogger();
        bool isDebugEnabled() const;
        void log(LogLevelEnum l, std::ostringstream& os, const char* filename, int lineno) const;
        std::ostringstream& getStream()
        { return os_; }
};

void Logger::log(LogLevelEnum l, std::ostringstream& os, const char* filename, int lineno) const
{
        std::cout << logLevelEnumToString(l) << "\t(" << fileName << ": " << lineno << ")\t- " << os.str();
        os.str("");
}

#define LOG_common(level, cptext) do {\
        utility::Logger::getLogger()->getStream() << cptext; \
        utility::Logger::getLogger()->log(utility::level, utility::Logger::getLogger()->getStream(), __FILE__, __LINE__);  \
} while(0); 

enum LogLevelEnum {
        DEBUG_LOG_LEVEL,
        INFO_LOG_LEVEL,
        WARN_LOG_LEVEL,
        ERROR_LOG_LEVEL,
        NOTICE_LOG_LEVEL,
        FATAL_LOG_LEVEL
};

#define LOG_DEBUG(cptext)    LOG_common(DEBUG_LOG_LEVEL, cptext)
#define LOG_INFO(cptext)     LOG_common(INFO_LOG_LEVEL , cptext)
#define LOG_WARN(cptext)     LOG_common(WARN_LOG_LEVEL , cptext)
#define LOG_ERROR(cptext)    LOG_common(ERROR_LOG_LEVEL, cptext)
#define LOG_NOTICE(cptext)   LOG_common(NOTICE_LOG_LEVEL, cptext)
#define LOG_FATAL(cptext)    LOG_common(FATAL_LOG_LEVEL, cptext)

const char* logLevelEnumToString(LogLevelEnum m)
{
        switch(m)
        {
                case DEBUG_LOG_LEVEL:
                        return "DEBUG";
                case INFO_LOG_LEVEL:
                        return "INFO";
                case WARN_LOG_LEVEL:
                        return "WARN";
                case NOTICE_LOG_LEVEL:
                        return "NOTICE";
                case ERROR_LOG_LEVEL:
                        return "ERROR";
                case FATAL_LOG_LEVEL:
                        return "FATAL";
                default:
                        CP_MSG_ASSERT(false, CP_TEXT("invalid value of LogLevelEnum"));
                        return 0;
        }
}

回答4:

You have to override operator<<(), but you even don't have to subclass std::cout. You may also create a new object or use existing objects like that.

回答5:

For logging purposes I use something like

#define LOG(x) \
  cout << __FUNCTION__ << x << endl

// ...
LOG("My message with number " << number << " and some more");

The problem with your approach is (as Mykola Golybyew explained) that FUNCTION is processed at compile time and would therefore always print the same name with a non-preprocessor solution.

If it's only for adding endl to your messages, you could try something like:

class MyLine {
public:
  bool written;
  std::ostream& stream;
  MyLine(const MyLine& _line) : stream(_line.stream), written(false) { }
  MyLine(std::ostream& _stream) : stream(_stream), written(false) { }
  ~MyLine() { if (!written) stream << "End of Message" << std::endl; }
};

template <class T> MyLine operator<<(MyLine& line, const T& _val) {
  line.stream << _val;
  line.written = true;
  return line;
}

class MyStream {
public:
  std::ostream& parentStream;
  MyStream(std::ostream& _parentStream) : parentStream(_parentStream) { }
  MyLine getLine() { return MyLine(parentStream); }
};

template <class T> MyLine operator<<(MyStream& stream, const T& _val) {
  return (stream.getLine() << _val);
}

int main()
{
      MyStream stream(std::cout);
      stream << "Hello " << 13 << " some more data";
      stream << "This is in the next line " << " 1 ";
    return 0;
}

Note, that it's important not to return references from the operator functions. Since the MyLine should only exist as a temporary (for its destructor triggers the writing of the endl), the first object (returned by the getLine() function in MyStream) would be destructed before the second operator<< is called. Therefore the MyLine object is copied in each operator<< creating a new one. The last object gets destructed without being written to and writed the end of the message in its destructor.

Just try it out in the debugger to understand whats going on...

回答6:

You could also override the operator. It will allow you to call another function or prefix/suffix anything that's going to leave the output buffer with whatever you wish: In your case, you'd have it output a specific string.

来源：https://stackoverflow.com/questions/674060/customize-cout

标签

c++

stl

iostream