CODEWITHSUNDEEP
c++partial-specializationc++-conceptsc++20
Mohit
Mohit

Reputation: 1295

Partial specialization using Concepts

I was just reading the examples of C++20 Concepts. Now I am trying to create a function that will print out if the given type is a hash-table or not using concepts mixed with the partial-specialization. But unfortunately it doesn't work.

#include <iostream>
#include <string>

template<typename T>
concept Hashtable = requires(T a) {
    { std::hash<T>{}(a) } -> std::size_t;
};

struct Foo {};

template <typename T>
void Bar() {
    std::cout << "Type T is not a hashtable" << std::endl;
}

template <Hashtable T>
void Bar<T> {
    std::cout << "Type T is a hashtable" << std::endl;
}

int main()
{
    Bar<Foo>();
    Bar<std::string>();
}

I am using compiler version GCC HEAD 9.0.1, compiler flags are g++ prog.cc -Wall -Wextra -I/opt/wandbox/boost-1.69.0/gcc-head/include -std=gnu++2a "-fconcepts". It gives me following compiler error:

prog.cc:18:6: error: template-id 'Bar<T>' used as a declarator
   18 | void Bar<T> {
      |      ^~~~~~
prog.cc:18:6: error: variable or field 'Bar' declared void
prog.cc:19:54: error: expected '}' before ';' token
   19 |     std::cout << "Type T is a hashtable" << std::endl;
      |                                                      ^
prog.cc:18:13: note: to match this '{'
   18 | void Bar<T> {
      |             ^
prog.cc:20:1: error: expected declaration before '}' token
   20 | }
      | ^

Live Demo

But my expectations were :

Type T is not a hashtable
Type T is a hashtable

My Question

Is it possible to specialize using Concepts?

Upvotes: 7

Views: 4026

Answers (3)

Barry
Barry

Reputation: 303007

Function templates cannot be partially specialized (and never could be). Concepts don't change that rule.

However, function templates can be overloaded (and always could be). And concepts do make that easier:

template <typename T>
void Bar() {
    std::cout << "Type T is not a hashtable" << std::endl;
}

template <Hashtable T>
void Bar() {
    std::cout << "Type T is a hashtable" << std::endl;
}

int main()
{
    Bar<Foo>();           // calls the first Bar
    Bar<std::string>();   // calls the second Bar
}

We say the second Bar is more constrained than the first Bar.

Upvotes: 9

P.W
P.W

Reputation: 26800

Is it possible to specialize using concepts?

No, it is not possible to partially specialize concepts. As per the online reference on Constraints and concepts:

Explicit instantiations, explicit specializations, or partial specializations of concepts are not allowed (the meaning of the original definition of a constraint cannot be changed).

And as far as function templates are concerned, they can only be overloaded, not partially specialized.

Upvotes: 0

Mohit
Mohit

Reputation: 1295

I've replaced the function template specialization with structs specialization and my code is working fine. Look at the following code.

// This file is a "Hello, world!" in C++ language by GCC for wandbox.
#include <iostream>
#include <string>

template<typename T>
concept Hashtable = requires(T a) {
    { std::hash<T>{}(a) } -> std::size_t;
};

struct Foo {};

template <typename T>
struct Boo {
    static constexpr char value[] = "Type T is not a hashtable";
};

template <Hashtable HashT>
struct Boo<HashT> {
    static constexpr char value[] = "Type T is a hashtable";
};

template <typename T>
void Bar() {
    std::cout << Boo<T>::value << std::endl;
}

int main()
{
    Bar<int>();
    Bar<Foo>();
}

Upvotes: 6

Related Questions