如何在 C++ 中使用模板重构方法

Question

我需要（或者更好的是，我有机会）重构一些代码以使其更简洁。

我想使用一些模板，因为我认为这是一个很好的候选，以减少代码重复。

这是我的 hpp

class Monetary
{
 public:
  Monetary();
  Monetary(const rapidjson::Value& iMonetary);
  virtual ~Monetary();

  [...cut...]

 private:

  static void initMember(const rapidjson::Value& iMonetary, const char* iName, int& oMember);
  static void initMember(const rapidjson::Value& iMonetary, const char* iName, std::string& oMember);


 private:
  int _amount;
  int _decimal_place;
  std::string _currency;
  std::string _type;
};

这是 initMember 方法的实现：

static void Monetary::initMember(const rapidjson::Value& iMonetary, const char* iName, int& oMember)
{
  rapidjson::Value::ConstMemberIterator aIterator;
  aIterator = iMonetary.FindMember(iName);
  if ( aIterator != iMonetary.MemberEnd() &&
      aIterator->value.IsNumber() )
  {
    oMember = iMonetary[iName].GetInt();
  }
}

static void Monetary::initMember(const rapidjson::Value& iMonetary, const char* iName, std::string& oMember)
{
  rapidjson::Value::ConstMemberIterator aIterator;
  aIterator = iMonetary.FindMember(iName);
  if ( aIterator != iMonetary.MemberEnd() &&
      aIterator->value.IsNumber() )
  {
    oMember = iMonetary[iName].GetString();
  }
}

我正在考虑写类似的东西

template<typename T>
void Monetary::initMember(const rapidjson::Value& iMonetary, const char* iName, T& oMember)
{
  rapidjson::Value::ConstMemberIterator aIterator;
  aIterator = iMonetary.FindMember(iName);
  if (aIterator == iMonetary.MemberEnd())
  {
    return;
    //throw monetaryException
  }
  assignFromValue(iMonetary[iName], oMember);
}
template<>
void Monetary::assignFromValue<int>(const rapidjson::Value& iValue, int& oMember)
{
  if (!iValue.IsNumber())
  {
    return;
    //throw monetaryException
  }
  oMember = iValue.GetInt();
}
template<>
void Monetary::assignFromValue<std::string>(const rapidjson::Value& iValue, std::string& oMember)
{
  if (!iValue.IsString())
  {
    return;
    //throw monetaryException
  }
  oMember = iValue.GetString();
}

有什么更简洁的方法吗？

Answer 1

我的建议：

1. 您不需要创建assignFromValue 成员函数。如果您可以使用非成员函数来实现功能，那么您应该更喜欢非成员函数。请参阅How Non-Member Functions Improve Encapsulation 和How Non-Member Functions Improve Encapsulation。

2. 您不需要制作assignFromValue 函数模板。它们可以是简单的重载。

---

void assignFromValue(const rapidjson::Value& iValue,
                     int& oMember)
{
  if (!iValue.IsNumber())
  {
    return;
    //throw monetaryException
  }
  oMember = iValue.GetInt();
}

void assignFromValue(const rapidjson::Value& iValue,
                     std::string& oMember)
{
  if (!iValue.IsString())
  {
    return;
    //throw monetaryException
  }
  oMember = iValue.GetString();
}

Answer 2

我想我会用标签调度的转换器对象来做：

#include <string>
#include <type_traits>
#include <stdexcept>

// simulate your value class
struct Value
{
    bool IsNumber() const;
    bool IsString() const;
    std::string getString() const;
    int getInt() const;
};


// a tag type for easy tag dispatching
template<class Type> struct tag {};


// a converter object contains all rules and conversion sequences
struct ValueConverter
{
    std::string operator()(tag<std::string>, const Value& v) const
    {
        if (not v.IsString()) throw std::invalid_argument("not a string");
        return v.getString();
    }

    int operator()(tag<int>, const Value& v) const
    {
        if (not v.IsNumber()) throw std::invalid_argument("not a number");
        return v.getInt();
    }
};

// write the member once
template<class Target>
void initMember(const Value& iMonetary, const char* iName, Target& oMember)
{
    using target_type = std::decay_t<Target>;
    auto converter = ValueConverter();
    oMember = converter(tag<target_type>(), iMonetary);
}