有时我会编写一个类(T说)并尝试覆盖std :: ostream&运算符<
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
}
我敢肯定,explicit关键字会阻止转换构造函数的情况,但是它确实会像这样,因为当我执行类似操作时
std::cout << config_ << std::endl;
它输出类似:0x100588140。但是然后我在另一堂课中做同样的事情,就像下面的课,一切都很好。
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
};
谢谢你的帮助。
编辑:
为了解决这个问题,我现在将以下内容添加到所有类中:
#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
最佳答案
尝试将签名更改为const
指针:
std::ostream& operator<<(std::ostream& output, const ConfigFile* c);