我已经创建了一个温度转换类,我决定创建一个具有如下结构的转换表:
temperature | conversion formula
出于这个原因,我创建了一个std :: map,其中键作为温度指数,然后是公式.在这里你可以看到实现:
enum class TempType {Celsius,Fahrenheit,Kelvin,Rankine,Delisle,Newton,Reaumur,Romer}; class Temperature { private: double value; //value of the temperature TempType kind; //Celsius,Fahrenheit... //conversion tables std::map<TempType,std::function<double(double)>> fromCelsius = { { TempType::Fahrenheit,[](double x) { return x * (9/5) + 32; } },{ TempType::Kelvin,[](double x) { return x + 273.15; } },{ TempType::Rankine,[](double x) { return (x + 273.15) * (9/5); } },{ TempType::Delisle,[](double x) { return (100 - x) * (3/2); } },{ TempType::Reaumur,[](double x) { return x * (33/100); } },{ TempType::Newton,[](double x) { return x * (4/5); } },{ TempType::Romer,[](double x) { return x * (21/40) + 7.5; } } }; std::map<TempType,std::function<double(double)>> fromFahrenheit = { { TempType::Celsius,[](double x) { return (x - 32) * (5/9); } },[](double x) { return (x + 459.67) * (5/9); } },[](double x) { return x + 459.67; } },[](double x) { return (212 - x) * (5/6); } },[](double x) { return (x - 32) * (11/60); } },[](double x) { return (x - 32) * (4/9); } },[](double x) { return (x - 32) * (7/24) + 7.5; } } }; std::map<TempType,std::function<double(double)>> fromKelvin = { { TempType::Celsius,[](double x) { return x - 273.15; } },{ TempType::Fahrenheit,[](double x) { return x * (9/5) - 459.67; } },[](double x) { return x * (9/5); } },[](double x) { return (373.15 - x) * (3/2); } },[](double x) { return (x - 273.15) * (33/100); } },[](double x) { return (x - 273.15) * (4/5); } },[](double x) { return (x - 273.15) * (21/40) + 7.5; } } }; std::map<TempType,std::function<double(double)>> fromRankine = { { TempType::Celsius,[](double x) { return (x - 491.67) * (5/9); } },[](double x) { return x - 459.67; } },[](double x) { return x * (5/9); } },[](double x) { return (671.67 - x) * (5/6); } },[](double x) { return (x - 491.67) * (11/60); } },[](double x) { return (x - 491.67) * (4/9); } },[](double x) { return (x - 491.67) * (7/24) + 7.5; } } }; std::map<TempType,std::function<double(double)>> fromDelisle = { { TempType::Celsius,[](double x) { return 100 - x * (5/9); } },[](double x) { return 212 - x * (6/5); } },[](double x) { return 373.15 - x * (2/3); } },[](double x) { return 671.67 - x * (6/5); } },[](double x) { return 33 - x * (11/50); } },[](double x) { return 80 - x * (8/15); } },[](double x) { return 60 - x * (7/20); } } }; std::map<TempType,std::function<double(double)>> fromNewton = { { TempType::Celsius,[](double x) { return x * (100/33); } },[](double x) { return x * (60/11) + 32; } },[](double x) { return x * (100/33) + 273.15; } },[](double x) { return x * (60/11) + 491.67; } },[](double x) { return (33 - x) * (50/11); } },[](double x) { return x * (80/33); } },[](double x) { return x * (35/22) + 7.5; } } }; std::map<TempType,std::function<double(double)>> fromReamur = { { TempType::Celsius,[](double x) { return x * (5/4); } },[](double x) { return x * (9/4) + 32; } },[](double x) { return x * (5/4) + 273.15; } },[](double x) { return x * (9/4) + 491.67; } },[](double x) { return (80 - x) * (15/8); } },[](double x) { return x * (33/80); } },[](double x) { return x * (21/32) + 7.5; } } }; std::map<TempType,std::function<double(double)>> fromRomer = { { TempType::Celsius,[](double x) { return (x - 7.5) * (40/21); } },[](double x) { return (x - 7.5) * (24/7) + 32; } },[](double x) { return (x - 7.5) * (40/21) + 273.15; } },[](double x) { return (x - 7.5) * (24/7) + 491.67; } },[](double x) { return (60 - x) * (20/7); } },[](double x) { return (x - 7.5) * (22/35); } },[](double x) { return (x - 7.5) * (32/21); } } }; public: } #endif // TEMPERATURE_H
现在,如果我想从温度对象转换为另一个,我使用此代码:
Temperature x(20,TempType::Celsius); //20 °C double s = x.convertTo(TempType::Fahrenheit); //convert 20 °c to 68 °F
转换代码是这样的:
double Temperature::convertTo(const TempType& temperature) const { if (temperature == kind) { return value; } double result = -1; switch (temperature) { case TempType::Celsius: result = fromCelsius.at(kind)(value); break; case TempType::Fahrenheit: result = fromFahrenheit.at(kind)(value); break; case TempType::Kelvin: result = fromKelvin.at(kind)(value); break; case TempType::Rankine: result = fromRankine.at(kind)(value); break; case TempType::Delisle: result = fromDelisle.at(kind)(value); break; case TempType::Newton: result = fromNewton.at(kind)(value); break; case TempType::Reaumur: result = fromReamur.at(kind)(value); break; case TempType::Romer: result = fromRomer.at(kind)(value); break; default: break; } return result; }
我使用地图的键(TempType),然后我将参数传递给lambda.返回的结果总是0.任何想法?
我正在使用QTCreator和mingw.
解决方法
在C 4/5中,21 / 40,5 / 6等都是零,因为除法是以整数运算执行的.
你需要写4./5.