Template member function specialization in a template class
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;
}
}
};
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)