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Home/ Questions/Q 6756965
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T13:35:20+00:00 2026-05-26T13:35:20+00:00

I have following code: class TR_AgentInfo : public tuple< long long, //AgentId string, //AgentIp

  • 0

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 ?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T13:35:20+00:00Added an answer on May 26, 2026 at 1:35 pm

    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.

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