偶尔我会写一个类(T说)并试图覆盖std :: ostream& operator<<(std :: ostream&,const T&)但它不适用于某些类。这是一个(简化)类的例子,它对我不起作用。
class ConfigFile {
public:
explicit ConfigFile(const std::string& filename);
virtual ~ConfigFile();
bool saveToDisk() const;
bool loadFromDisk();
std::string getSetting(const std::string& setting, const std::string& section="Misc") const;
void setSetting(std::string value, const std::string& name, const std::string& section ="Misc", bool updateDisk = false);
inline const SettingSectionMap& getSettingMap() const {
return mSettingMap;
}
private:
std::string mSettingFileName;
SettingSectionMap mSettingMap;
#if defined(_DEBUG) || defined(DEBUG)
public:
friend std::ostream& operator<<(std::ostream& output, const ConfigFile& c) {
output << “Output the settings map here”;
return output;
}
#endif
}
我很确定显式关键字会阻止转换构造函数的情况,但它肯定会像它一样,因为当我做类似的事情时
std::cout << config_ << std::endl;
它输出的内容如下:0x100588140。但是后来我在另一个类中执行相同的操作,如下所示,一切正常。
class Stats {
Stats() {};
#if defined(_DEBUG) || defined(DEBUG)
friend std::ostream& operator<<(std::ostream& output, const Stats& p) {
output << "FPS Stats: " << p.lastFPS_ << ", " << p.avgFPS_ << ", " << p.bestFPS_ << ", " << p.worstFPS_ << " (Last/Average/Best/Worst)";
return output;
};
#endif
};
感谢您的帮助。
修改 为了解决这个问题,我现在将以下内容添加到我的所有类中:
#if defined(_DEBUG) || defined(DEBUG)
public:
friend std::ostream& operator<<(std::ostream& output, const ConfigFile& c);
friend std::ostream& operator<<(std::ostream& output, ConfigFile* c) {
output << *c;
return output;
}
#endif
答案 0 :(得分:0)
尝试将签名更改为const
指针:
std::ostream& operator<<(std::ostream& output, const ConfigFile* c);