Template member function specialization in a template class

Question

I have a template class and a member function print() to print the data.

template<typename T>
class A
{
public:
   T data;
   void print(void) 
   { 
      std::cout << data << std::endl; 
   }
   // other functions ...
};

Then, I want to either print scalar data or vector data, so I give a specialized definition and get a compiler error.

template<typename T>
void A<std::vector<T>>::print(void) // template argument list error
{
   for (const auto& d : data) 
   {
      std::cout << d << std::endl;
   }
}

Question: Why does this member function specialization get an error? What is the correct way to define a print function for a vector?

Solution 1: I have tested the following definition.

template<typename T>
class A<std::vector<T>>
{
public:
   std::vector<T> data;
   void print(void) {   // OK
      // ... 
   } 
}

This one worked, but I have to copy the other member functions into this specialized class.

---

EDIT:

Solution 2: To prevent copy all the other member functions, I define a base class containing the common member functions and inherit from the base class:

template<typename T>
class Base
{
public:
   T data;
   // other functions ...
};

template<typename T>
class A : public Base<T>
{
public:
   void print(void) 
   {
      std::cout << this->data << std::endl;
   }
};

template<typename T>
class A<std::vector<T>> : public Base<std::vector<T>>
{
public:
   void print(void) 
   {
      for (const auto& d : this->data)
      {
         std::cout << d << std::endl;
      }
   }
};

This solution works well. Are there some better or more conventional solutions?

Answer 1

Why does this member function specialization get error?

When you instantiate the template class A for example A<std::vector<int>>, the template parameter T is equal to std::vector<int>, not std::vector<T>, and this a specialization case of the function. Unfortunately this can not be done with member functions as mentioned in the comments.

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Are there some better solutions?

Yes; In c++17 you could use if constexpr with a trait to check the std::vector, like this.

#include <type_traits> // std::false_type, std::true_type
#include <vector>

// traits for checking wether T is a type of std::vector<>
template<typename T> struct is_std_vector final : std::false_type {};
template<typename... T> struct is_std_vector<std::vector<T...>> final : std::true_type {};

template<typename T>
class A /* final */
{
    T mData;

public:  
    // ...constructor  

    void print() const /* noexcept */
    {
        if constexpr (is_std_vector<T>::value) // when T == `std::vector<>`
        {
            for (const auto element : mData)
                std::cout << element << "\n";
        }
        else // for types other than `std::vector<>` 
        {
            std::cout << mData << std::endl;
        }
    }
};

(See Live Online)

This way you keep only one template class and the print() will instantiate the appropriate part according to the template type T at compile time.

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If you don not have access to C++17, other option is to SFINAE the members(Since c++11).

#include <type_traits> // std::false_type, std::true_type, std::enbale_if
#include <vector>

// traits for checking wether T is a type of std::vector<>
template<typename T> struct is_std_vector final : std::false_type {};
template<typename... T> struct is_std_vector<std::vector<T...>> final : std::true_type {};

template<typename T>
class A /* final */
{
    T mData;

public:  
    // ...constructor  

    template<typename Type = T> // when T == `std::vector<>`
    auto print() const -> typename std::enable_if<is_std_vector<Type>::value>::type
    {
        for (const auto element : mData)
                std::cout << element << "\n";
    }

    template<typename Type = T> // for types other than `std::vector<>`
    auto print() const -> typename std::enable_if<!is_std_vector<Type>::value>::type
    {
        std::cout << mData << std::endl;
    }
};

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What if I have more other data types like self-define vector classes or matrices? Do I have to define many is_xx_vector?

You can check the type is a specialization of the provided one like as follows. This way you can avoid providing many traits for each type. The is_specialization is basically inspired from this post

#include <type_traits> // std::false_type, std::true_type
#include <vector>

// custom MyVector (An example)
template<typename T> struct MyVector {};

template<typename Test, template<typename...> class ClassType>
struct is_specialization final : std::false_type {};

template<template<typename...> class ClassType, typename... Args>
struct is_specialization<ClassType<Args...>, ClassType> final : std::true_type {};

And the print function could be in c++17:

void print() const /* noexcept */
{
   if constexpr (is_specialization<T, std::vector>::value)// when T == `std::vector<>`
   {
      for (const auto element : mData)
         std::cout << element << "\n";
   }
   else if constexpr (is_specialization<T, ::MyVector>::value)  // custom `MyVector`
   {
      std::cout << "MyVector\n";
   }
   else  // for types other than `std::vector<>` and custom `MyVector`
   {
      std::cout << mData << std::endl;
   }
}

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