tuple_size and an inhereted class from tuple?

问题:

I have following code:

class TR_AgentInfo : public tuple<
                              long long,         //AgentId
                              string,         //AgentIp
                              >
{
public:
TR_AgentInfo() {}
TR_AgentInfo(
  const long long& AgentId,
  const string& AgentIp,
   )
{
    get<0>(*this) = AgentId;
    get<1>(*this) = AgentIp;
}

long long getAgentId() const { return get<0>(*this); }
void setAgentId(const long long& AgentId) { get<0>(*this) = AgentId; }

string getAgentIp() const { return get<1>(*this); }
void setAgentIp(const string& AgentIp) { get<1>(*this) = AgentIp; }
};

Now I want to use this code:

int count = tuple_size<TR_AgentInfo>::value;

but gcc give this error:

error: incomplete type std::tuple_size<TR_AgentInfo> used in nested name specifier

now what can I do ?

回答1:

If you just want your one class to work with std::tuple_size, you can simply provide a specialization:

namespace std
{
  template<> struct tuple_size<TR_AgentInfo>
  {
    static const size_t value = 2;

    // alternatively, `tuple_size<tuple<long long, string>>::value`
    // or even better, `tuple_size<TR_AgentInfo::tuple_type>::value`, #1
  };
}

You are expressly allowed to add specializations to the namespace std, precisely for situations like yours.

If your actual class is itself templated, you can simply replace 2 by an appropriate construction. For example, for suggestion #1 you could add a typedef for tuple_type to your class. There are many ways to skin this cat.

回答2:

If you have an instance of TR_AgentInfo kicking around, you can use function template type deduction to make it friendlier, like so:

template <typename... T>
std::size_t get_tuple_size(std::tuple<T...> const &)
{
  return sizeof...(T);
}

and call

TR_AgentInfo info;
int count = get_tuple_size(info);

otherwise you need to use proper metaprogramming to get at the base class, but facilities are available.