Template member function specialization in a template class - c++

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.
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.
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;
}
};
(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 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;
}
}
(See Live Online)

Answer 2

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;
}
}
};

Answer 3

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

Answer 4

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;
}
};