CODEWITHSUNDEEP
c++c++11templatessfinae
Alex Aparin
Alex Aparin

Reputation: 4522

How to use sfinae to check, whether type has operator ()?

I have following code:

template <typename T>
struct function_traits
{
    typedef decltype(&T::operator()) test_type;
    typedef std::true_type res_type;
};

template <typename T>
struct function_traits
{
    typedef std::false_type res_type;
};

In other words, I want to know whether type has operator (). I thought that I can use SFINAE way to do this. However compiler tells:

'function_traits' : class template has already defined.

What's wrong with such code?

P.S.: here is simple usage:

auto f1 = [](const int a){ std::cout << a << std::endl; };
function_traits<decltype(f1)>::res_type;
auto f2 = false;
function_traits<decltype(f2)>::res_type;

EDIT: I am using c++ 11 standard

Upvotes: 3

Views: 3469

Answers (2)

Oliv
Oliv

Reputation: 18081

Maybe using the void_t is much more simple:

template <typename T,class=void>
struct callable_without_args: std::false_type{};

template <typename T>
struct callable_without_args<T
          ,std::void_t<decltype(std::declval<T>()())>>
  :std::true_type{};

It could be even simpler if your compiler provided concepts:

template<T>
concept bool CallableWithoutArgs= requires(T&& a){
     (T&&)(a)();
};

Upvotes: 3

javaLover
javaLover

Reputation: 6425

Encouraged by a positive feedback, I will post a brief answer here.

The problem is that you cannot define a class twice, but you did this two times :-

template <typename T>
struct function_traits {  .... some code ..... }

C++ doesn't allow that.

To check whether a function is exist, there is already a question about it, you can modify it to support the operator().

Here is a demo, very-slightly modified from Nicola Bonelli's answer there. 

#include <iostream>

struct Hello
{
    void operator()() {  }
};

struct Generic {};    

// SFINAE test
template <typename T>
class has_parenthesis
{
    typedef char one;
    typedef long two;

    template <typename C> static one test( decltype(&C::operator()) ) ;
    template <typename C> static two test(...);    

public:
    enum { value = sizeof(test<T>(0)) == sizeof(char) };
};

int main(int argc, char *argv[])
{
    std::cout << has_parenthesis<Hello>::value << std::endl;    //print 1
    std::cout << has_parenthesis<Generic>::value << std::endl;  //print 0
    return 0;
}

I just knew a few minutes ago that it also works for operator().

Edit: I changed typeof to decltype as StoryTeller and LmTinyToon recommended. Thank.

Upvotes: 6

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