Template member function specialization in a template class

15
votes

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?

4
c++templatesc++17template-specializationmember-functions
Does this answer your question? c++ template partial specialization member function – G. Sliepen
I think the example in my question is not a partial specialization. – kfckfckf
Are there some better or more conventional solutions?: This could be opinion based answer. You should go for those which suits to your situation. If you have many members to do like this, a specialized class would be a better option as you could collectively see them in one class definition. – JeJo

4 Answers

10
votes

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.

Are there some better solutions?

Yes; In 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.

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

#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;
    }
};

(See Live Online)

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 :

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;
   }
}

(See Live Online)

5
votes

You need to implement a template class that uses a vector as template parameter. This worked for me.

template<typename T>
class A
{
public:
    T data;

    void print(void) {
        std::cout << "Data output" << std::endl;
    }
    // other functions ...
};

template <typename T>
class A<std::vector<T>>
{
public:
    std::vector<T> data;

    void print() {
        for (auto i : data) {
            std::cout << "Vector output" << std::endl;
        }
    }
};
3
votes

You could always use named tag dispatching to check if type provided by template user is vector.

A<std::vector<T>> notation won't work as you both try to take into account that T is type and vector of types which is contradicting with itself.

Below is code I used named tag dispatching as solution to your problem:

#include <iostream>
#include <vector>
#include <type_traits>

using namespace std;

template<typename T> struct is_vector : public std::false_type {};

template<typename T, typename A>
struct is_vector<std::vector<T, A>> : public std::true_type {};

template<typename T>
class A
{
public:
    T data;
    void print(std::true_type) {
        for (auto& a : data) { std::cout << a << std::endl; } 
    }
    void print(std::false_type) {
        std::cout << data << std::endl;
    }
    void print() {
        print(is_vector<T>{});
    }
};
    
int main()
{
    A<int> a;
    a.data = 1;
    a.print();
    A<std::vector<int>> b;
    b.data = { 1, 2 ,3 ,4 ,5 };
    b.print();
    return 0;
}

Succesfully compiled with https://www.onlinegdb.com/online_c++_compiler

Based on answer: Check at compile-time is a template type a vector

2
votes

You can dispatch printing to another member function (static or not). For example:

template<typename T>
class A {
public:
    T data;
    
    void print() const {
        print_impl(data);
    }
    
private:
    template<class S>
    static void print_impl(const S& data) {
        std::cout << data;
    }

    template<class S, class A>
    static void print_impl(const std::vector<S, A>& data) {
        for (const auto& d : data)
            std::cout << d;
    }
};