c++ - 如何简化这个冗余的 C++ 代码？

Question

现有的类有两个，一个是SrcField它返回具体类型值，另一个是 union DSTField ，它定义了相应的数据类型。

class SrcField
{
public:
    signed char     GetInt8();
    unsigned char   GetUInt8();
    short           GetInt16();
    unsigned short  GetUInt16();
    int             GetInt32();
    unsigned int    GetUInt32();
    float           GetFloat();
    double          GetDouble();
    bool            GetBool();
    DataType        GetType();
private:
    DataType m_type;
    DSTField m_data;
};

union DSTField
{
    signed char    m_int8;
    unsigned char  m_uint8;
    short          m_int16;
    unsigned short m_uint16;
    int            m_int32;
    unsigned int   m_uint32;
    float          m_float;
    double         m_double;
    bool           m_bool;
};

当我使用这两个类时，应用程序如下。这是非常多余的；有没有什么好的简化方法，比如模板、泛型编程等？

int main()
{
    SrcField sf;
    DSTField df;

    switch(sf.GetType())
    {
    case TYPE_INT8:
        df.m_int8 = sf.GetInt8();
        break;
    case TYPE_UINT8:
        df.m_uint8 = sf.GetUInt8();
        break;
    case TYPE_INT16:
        df.m_int16 = sf.GetInt16();
        break;
    case TYPE_UINT16:
        df.m_uint16 = sf.GetUInt16();
        break;
    case TYPE_INT32:
        df.m_int32 = sf.GetInt32();
        break;
    case TYPE_UINT32:
        df.m_uint32 = sf.GetUInt32();
        break;
    case TYPE_FLOAT:
        df.m_float = sf.GetFloat();
        break;
    case TYPE_DOUBLE:
        df.m_double = sf.GetDouble();
        break;
    case TYPE_BOOL:
        df.m_bool = sf.GetBool();
        break;
    default:
        break;
    }
}

Answer 1

使用 std::variant 您的代码将如下所示:

#include <iostream>
#include <variant>

typedef std::variant<
    signed char,
    unsigned char,
    short,
    unsigned short,
    int,
    unsigned int,
    float,
    double,
    bool
> SrcField, DSTField;

int main()
{
    SrcField sf(97.0f);
    DSTField df;

    df = sf;

    if(auto pval = std::get_if<float>(&df))
      std::cout << "variant value: " << *pval << '\n'; 
    else 
      std::cout << "failed to get value!" << '\n';
}

---

注意:由于是 c++17，对于以前的版本，我建议使用 boost::variant、boost::any 或 Any 类(例如我在我的项目中使用一个基于 this 的类)