Q:

Detect operator support with decltype/SFINAE

A (somewhat) outdated article explores ways to use decltype along with SFINAE to detect if a type supports certain operators, such as == or <.  
Here's example code to detect if a class supports the < operator:
template <class T>
struct supports_less_than
{
    static auto less_than_test(const T* t) -> decltype(*t < *t, char(0))
    { }

    static std::array<char, 2> less_than_test(...) { }

    static const bool value = (sizeof(less_than_test((T*)0)) == 1);
};

int main()
{
    std::cout << std::boolalpha << supports_less_than<std::string>::value << endl;
}

This outputs true, since of course std::string supports the < operator.  However, if I try to use it with a class that doesn't support the < operator, I get a compiler error:
error: no match for ‘operator<’ in ‘* t < * t’

So SFINAE is not working here.  I tried this on GCC 4.4 and GCC 4.6, and both exhibited the same behavior.  So, is it possible to use SFINAE in this manner to detect whether a type supports certain expressions?  

A:

In C++11 the shortest most general solution I found was this one:
#include <type_traits>

template<class T, class = decltype(std::declval<T>() < std::declval<T>() )> 
std::true_type  supports_less_than_test(const T&);
std::false_type supports_less_than_test(...);

template<class T> using supports_less_than = decltype(supports_less_than_test(std::declval<T>()));

#include<iostream>
struct random_type{};
int main(){
    std::cout << supports_less_than<double>::value << std::endl; // prints '1'
    std::cout << supports_less_than<int>::value << std::endl; // prints '1'
    std::cout << supports_less_than<random_type>::value << std::endl; // prints '0'
}

Works with g++ 4.8.1 and clang++ 3.3

A more general solution for arbitrary operators (UPDATE 2014)
There is a more general solution that exploits the fact that all built-in operators are also accessible (and posibly specialized) through STD operator wrappers, such as std::less (binary) or std::negate (unary). 
template<class F, class... T, typename = decltype(std::declval<F>()(std::declval<T>()...))> 
std::true_type  supports_test(const F&, const T&...);
std::false_type supports_test(...);

template<class> struct supports;
template<class F, class... T> struct supports<F(T...)> 
: decltype(supports_test(std::declval<F>(), std::declval<T>()...)){};

This can be used in a quite general way, especially in C++14, where type deduction is delayed to the operator wrapper call ("transparent operators").
For binary operators it can be used as: 
#include<iostream>
struct random_type{};
int main(){
    std::cout << supports<std::less<>(double, double)>::value << std::endl; // '1'
    std::cout << supports<std::less<>(int, int)>::value << std::endl; // '1'
    std::cout << supports<std::less<>(random_type, random_type)>::value << std::endl; // '0'
}

For unary operators:
#include<iostream>
struct random_type{};
int main(){
    std::cout << supports<std::negate<>(double)>::value << std::endl; // '1'
    std::cout << supports<std::negate<>(int)>::value << std::endl; // '1'
    std::cout << supports<std::negate<>(random_type)>::value << std::endl; // '0'
}

(With the C++11 standard library is a bit more complicated because there is no failure on instatiating decltype(std::less<random_type>()(...)) even if there is no operation defined for random_type, one can implement manually transparent operators in C++11, that are standard in C++14)
The syntax is quite smooth. I hope something like this is adopted in the standard.

Two extensions:
1) It works to detect raw-function applications: 
struct random_type{};
random_type fun(random_type x){return x;}
int main(){
    std::cout << supports<decltype(&fun)(double)>::value << std::endl; // '0'
    std::cout << supports<decltype(&fun)(int)>::value << std::endl; // '0'
    std::cout << supports<decltype(&fun)(random_type)>::value << std::endl; // '1'
}

2) It can additionally detect if the result is convertible/comparable to a certain type, in this case double < double is supported but a compile-time false will be returned because the result is not the specified one.
std::cout << supports<std::equal_to<>(std::result_of<std::less<>(double, double)>::type, random_type)>::value << std::endl; // '0'

Note: I just tried to compile the code with C++14 in http://melpon.org/wandbox/ and it didn't work. I think there is a problem with transparent operators (like std::less<>) in that implementation (clang++ 3.5 c++14), since when I implement my own less<> with automatic deduction it works well.

A:

You need to make your less_than_test function a template, since SFINAE stands for Substitution Failure Is Not An Error and there's no template function that can fail selection in your code.
template <class T>
struct supports_less_than
{
    template <class U>
    static auto less_than_test(const U* u) -> decltype(*u < *u, char(0))
    { }

    static std::array<char, 2> less_than_test(...) { }

    static const bool value = (sizeof(less_than_test((T*)0)) == 1);
};

int main()
{
    std::cout << std::boolalpha << supports_less_than<std::string>::value << endl;
}

A:

This is C++0x, we don't need sizeof-based tricks any more... ;-]
#include <type_traits>
#include <utility>

namespace supports
{
    namespace details
    {
        struct return_t { };
    }

    template<typename T>
    details::return_t operator <(T const&, T const&);

    template<typename T>
    struct less_than : std::integral_constant<
        bool,
        !std::is_same<
            decltype(std::declval<T const&>() < std::declval<T const&>()),
            details::return_t
        >::value
    > { };
}

(This is based on iammilind's answer, but doesn't require that T's operator< return-type be a different size than long long and doesn't require that T be default-constructable.)

