How to specialize template member function?

Question

I have the following template method:

struct MyStruct
{
  // ...
  template<typename T>
  void readField(std::istream& in, T& data)
  {
      read(in, data);
      data = ntohl(data);
  }
};

template<>
void MyStruct::readField<uint8_t>(std::istream& in, uint8_t& data)
{
    read(in, data);
}

But I get those strange linker errors:

/usr/lib/gcc/x86_64-redhat-linux/4.4.7/../../../../include/c++/4.4.7/exception:62: multiple definition of `void MyStruct::readField(std::basic_istream >&, unsigned char&)' ../Lib/obj/MyStruct.o:/usr/lib/gcc/x86_64-redhat-linux/4.4.7/../../../../include/c++/4.4.7/exception:62: first defined here collect2: ld returned 1 exit status make: *** [Lib] Error 1

How can I specialize this member function?

EDIT

This approach works:

struct MyStruct
{
  // ...
  template<typename T>
  void readField(std::istream& in, T& data)
  {
    read(in, data);
    data = ntohl(data);
  }

  void readField(std::istream& in, uint8_t& data)
  {
    read(in, data);
  } 
};

or with inlines or specializing it outside the class with inline

struct MyStruct
{
  // ...
  template<typename T>
  void readField(std::istream& in, T& data)
  {
      read(in, data);
      data = ntohl(data);
  }
};

template<>
inline void MyStruct::readField<uint8_t>(std::istream& in, uint8_t& data)
{
    read(in, data);
}

Function specialization is a regular function, not a template. As any other function, it must be declared in a header file and implemented in a source file; or else, implemented in a header file but marked inline. — Igor Tandetnik
@IgorTandetnik This approach works. You can submit it as an answer and I can accept it. Although I can define this method in the header file without inline. Why is that so? — Patryk
I can define this method in the header file without inline Clearly you cannot, or you woudln't be here asking questions. You get a linker error when you try, don't you? Perhaps I misunderstand what you are trying to say. — Igor Tandetnik
@IgorTandetnik Ok so what I meant was that I can define the method (without any templates/specialization) in the header file inside the class without inline. What I can do is as you mentioned specialize it with inline but only outside the class definition. Thanks a lot — Patryk
An in-class definition of a member function is implicitly inline. — Igor Tandetnik

RobClucas RobClucas · Accepted Answer · 2016-01-26T15:43:45

As Igor mentioned, you can implement the generic version in the header file, and then the specialization in the cpp file, for example:

// MyStruct.h

struct MyStruct {
  // ...
  template <typename T>
  void readField(std::istream& in, T& data) {
    read(in, data);
    data = ntohl(data);
  }
};

Then in the cpp file you can implement the specialization, for example:

// MyStruct.cpp

template <>
void MyStruct::readField<uint8_t>(std::istream& in, uint8_t& data) {
  read(in, data);
}

Update: After reading the comments, the specialization can also be in the same header file as the primary template, but not within the struct, for example (I verified this by compiling and running a similar example without error):

// MyStruct.h

struct MyStruct {
  // ...
  template <typename T>
  void readField(std::istream& in, T& data) {
    read(in, data);
    data = ntohl(data);
  }
};  

template <>
inline void MyStruct::readField<uint8_t>(std::istream& in, uint8_t& data) {
  read(in, data);
}

// End MyStruct.h

How to specialize template member function?

2 Answers