Question

easylogging ++代码定义了一个使其易于使用的宏：

LOG(logLevel) << "This mimics std::cout syntax.  " << 1 << " + " << 1 << " = " << 2;

我想为easylogging ++创建一个包装类。我可以轻松地创建一个带有两个参数的函数来包装上面的行。但是，是否可以在包装类中模仿这种语法？例如：

Logger logger;
logger(logLevel) << "Line " << 1 << " of log text.";

我知道我可以轻松地重载插入操作符，但这仍然让我不得不每次都写另一个函数来设置日志级别。

更新：

感谢Starl1ght的回答，我得到了这个工作。我想我会分享以防其他人有类似的需求。

我创建了两个重载。一个用于（），另一个用于＆lt;＆lt;。

Logger &operator()(logLevelT logLevel) {
  mLogLevel = logLevel;
  return *this;
}


template <typename T>
Logger &operator<<(T const &value) {
  LOG(mLogLevel) << value;
  return *this;
}

更新2：

我想再次更新这篇文章以提供我的推理并展示我的最终解决方案。

我的理由是我的项目是抽象的证明。我试图证明日志库（以及许多其他东西）可以从软件的核心功能中抽象出来。这也使软件组件模块化。这样我就可以换掉easylogging ++库而不会丢失语法，因为它是在模块接口中实现的。

我的上一次更新没有提到我如何克服插入链的障碍，所以我想发一个例子来说明我是如何做到的。以下代码是如何为类实现std :: cout like语法的简化示例。

#include <iostream>         // For cout
#include <string>           // For strings
#include <sstream>          // For ostringstream


enum logLevelT {
    INFO_LEVEL,
    WARNING_LEVEL,
    ERROR_LEVEL,
    FATAL_LEVEL
};


class Logger {
private:
    std::string logName;

public:
    Logger(std::string nameOfLog, std::string pathToLogFile) {
        logName = nameOfLog;

        //TODO Configure your logging library and instantiate
        //     an instance if applicable.
    }


    ~Logger(){}


    // LogInputStream is instantiated as a temporary object.  It is used
    //  to build the log entry stream.  It writes the completed stream
    //  in the destructor as the object goes out of scope automatically.
    struct LogInputStream {
        LogInputStream(logLevelT logLevel, std::string nameOfLog) {
            currentLogLevel = logLevel;
            currentLogName = nameOfLog;
        }


        // Copy Constructor
        LogInputStream(LogInputStream &lis) {
            currentLogLevel = lis.currentLogLevel;
            currentLogName = lis.currentLogName;
            logEntryStream.str(lis.logEntryStream.str());
        }


        // Destructor that writes the log entry stream to the log as the
        //  LogInputStream object goes out of scope.
        ~LogInputStream() {
            std::cout << "Logger: " << currentLogName
                      << "   Level: " << currentLogLevel
                      << "   logEntryStream = " << logEntryStream.str()
                      << std::endl;

            //TODO Make a log call to your logging library.  You have your log level
            //     and a completed log entry stream.
        }


        // Overloaded insertion operator that adds the given parameter
        //  to the log entry stream.
        template <typename T>
        LogInputStream &operator<<(T const &value) {
            logEntryStream << value;
            return *this;
        }


        std::string currentLogName;
        logLevelT currentLogLevel;
        std::ostringstream logEntryStream;
    };


    // Overloaded function call operator for providing the log level
    Logger::LogInputStream operator()(logLevelT logLevel) {
        LogInputStream logInputStream(logLevel, logName);

        return logInputStream;
    }


    // Overloaded insertion operator that is used if the overloaded
    //  function call operator is not used.
    template <typename T>
    Logger::LogInputStream operator<<(T const &value) {
        LogInputStream logInputStream(INFO_LEVEL, logName);

        logInputStream << value;

        return logInputStream;
    }
};



int main(int argc, char *argv[]) {

    Logger logger1 = Logger("Logger1", "/path/to/log.log");
    Logger logger2 = Logger("Logger2", "/path/to/log.log");

    logger1(INFO_LEVEL) << "This is the " << 1 << "st test";

    logger2(ERROR_LEVEL) << "This is the " << 2 << "nd test";

    logger2 << "This is the " << 3 << "rd test";

    return 0;
}

我觉得我可以用命名和评论做得更好，但是时间紧迫。我绝对愿意接受任何评论或批评。

Answer 1

您必须重载operator()因此，它将设置内部日志级别并返回*this作为类型Logger&，因此，重载operator<<将使用必要的日志返回引用-level set。

这样的事情：

Logger& Logger::operator()(LogLevel level) {
    // set internal log level
    return *this;
}

是否有可能通过链接重载插入运算符来创建一个模仿std :: cout语法的C ++类，就像使用easylogging ++一样？

1 个答案: