泛型模板，用于检查类型是否支持某些“特性”:

#include <type_traits>

template <template <typename> class TypeChecker, typename Type>
struct is_supported
{
    // these structs are used to recognize which version
    // of the two functions was chosen during overload resolution
    struct supported {};
    struct not_supported {};

    // this overload of chk will be ignored by SFINAE principle
    // if TypeChecker<Type_> is invalid type
    template <typename Type_>
    static supported chk(typename std::decay<TypeChecker<Type_>>::type *);

    // ellipsis has the lowest conversion rank, so this overload will be
    // chosen during overload resolution only if the template overload above is ignored
    template <typename Type_>
    static not_supported chk(...);

    // if the template overload of chk is chosen during
    // overload resolution then the feature is supported
    // if the ellipses overload is chosen the the feature is not supported
    static constexpr bool value = std::is_same<decltype(chk<Type>(nullptr)),supported>::value;
};

检查方法foo是否与signature double兼容的模板(const char*)

// if T doesn't have foo method with the signature that allows to compile the bellow
// expression then instantiating this template is Substitution Failure (SF)
// which Is Not An Error (INAE) if this happens during overload resolution
template <typename T>
using has_foo = decltype(double(std::declval<T>().foo(std::declval<const char*>())));

例子

// types that support has_foo
struct struct1 { double foo(const char*); };            // exact signature match
struct struct2 { int    foo(const std::string &str); }; // compatible signature
struct struct3 { float  foo(...); };                    // compatible ellipsis signature
struct struct4 { template <typename T>
                 int    foo(T t); };                    // compatible template signature

// types that do not support has_foo
struct struct5 { void        foo(const char*); }; // returns void
struct struct6 { std::string foo(const char*); }; // std::string can't be converted to double
struct struct7 { double      foo(      int *); }; // const char* can't be converted to int*
struct struct8 { double      bar(const char*); }; // there is no foo method

int main()
{
    std::cout << std::boolalpha;

    std::cout << is_supported<has_foo, int    >::value << std::endl; // false
    std::cout << is_supported<has_foo, double >::value << std::endl; // false

    std::cout << is_supported<has_foo, struct1>::value << std::endl; // true
    std::cout << is_supported<has_foo, struct2>::value << std::endl; // true
    std::cout << is_supported<has_foo, struct3>::value << std::endl; // true
    std::cout << is_supported<has_foo, struct4>::value << std::endl; // true

    std::cout << is_supported<has_foo, struct5>::value << std::endl; // false
    std::cout << is_supported<has_foo, struct6>::value << std::endl; // false
    std::cout << is_supported<has_foo, struct7>::value << std::endl; // false
    std::cout << is_supported<has_foo, struct8>::value << std::endl; // false

    return 0;
}

http://coliru.stacked-crooked.com/a/83c6a631ed42cea4

这就是类型特征存在的意义。不幸的是，它们必须手动定义。在你的情况下，想象一下:

template <typename T>
struct response_trait {
    static bool const has_tostring = false;
};

template <>
struct response_trait<your_type_with_tostring> {
    static bool const has_tostring = true;
}

是的，使用SFINAE您可以检查给定的类是否提供了特定的方法。下面是工作代码:

#include <iostream>

struct Hello
{
    int helloworld() { return 0; }
};

struct Generic {};    

// SFINAE test
template <typename T>
class has_helloworld
{
    typedef char one;
    struct two { char x[2]; };

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

public:
    enum { value = sizeof(test<T>(0)) == sizeof(char) };
};
    
int main(int argc, char *argv[])
{
    std::cout << has_helloworld<Hello>::value << std::endl;
    std::cout << has_helloworld<Generic>::value << std::endl;
    return 0;
}

我刚刚用Linux和gcc 4.1/4.3测试了它。我不知道它是否可以移植到运行不同编译器的其他平台。

虽然这个问题是两年前的事了，但我敢补充我的答案。希望它能澄清之前无可争议的优秀解决方案。我采纳了Nicola Bonelli和Johannes Schaub非常有用的答案，并将它们合并到一个解决方案中，恕我之言，这个解决方案更易于阅读，更清晰，不需要扩展类型:

template <class Type>
class TypeHasToString
{
    // This type won't compile if the second template parameter isn't of type T,
    // so I can put a function pointer type in the first parameter and the function
    // itself in the second thus checking that the function has a specific signature.
    template <typename T, T> struct TypeCheck;

    typedef char Yes;
    typedef long No;

    // A helper struct to hold the declaration of the function pointer.
    // Change it if the function signature changes.
    template <typename T> struct ToString
    {
        typedef void (T::*fptr)();
    };

    template <typename T> static Yes HasToString(TypeCheck< typename ToString<T>::fptr, &T::toString >*);
    template <typename T> static No  HasToString(...);

public:
    static bool const value = (sizeof(HasToString<Type>(0)) == sizeof(Yes));
};

我用gcc 4.1.2检查了它。 这主要归功于尼古拉·博内利和约翰内斯·绍布，如果我的回答对你有帮助，请给他们投票:)

这是c++ 17中另一种实现方法(灵感来自boost:hana)。

该解决方案不需要has_something<T> SFINAE类型trait类。

解决方案

////////////////////////////////////////////
// has_member implementation
////////////////////////////////////////////

#include <type_traits>

template<typename T, typename F>
constexpr auto has_member_impl(F&& f) -> decltype(f(std::declval<T>()), true)
{
  return true;
}

template<typename>
constexpr bool has_member_impl(...) { return false; }

#define has_member(T, EXPR) \
 has_member_impl<T>( [](auto&& obj)->decltype(obj.EXPR){} )

Test

////////////////////////////////////////////
// Test
////////////////////////////////////////////

#include <iostream>
#include <string>

struct Example {
    int Foo;
    void Bar() {}
    std::string toString() { return "Hello from Example::toString()!"; }
};

struct Example2 {
    int X;
};

template<class T>
std::string optionalToString(T* obj)
{
    if constexpr(has_member(T, toString()))
        return obj->toString();
    else
        return "toString not defined";
}

int main() {
    static_assert(has_member(Example, Foo), 
                  "Example class must have Foo member");
    static_assert(has_member(Example, Bar()), 
                  "Example class must have Bar() member function");
    static_assert(!has_member(Example, ZFoo), 
                  "Example class must not have ZFoo member.");
    static_assert(!has_member(Example, ZBar()), 
                  "Example class must not have ZBar() member function");

    Example e1;
    Example2 e2;

    std::cout << "e1: " << optionalToString(&e1) << "\n";
    std::cout << "e1: " << optionalToString(&e2) << "\n";
}

你可以跳过c++ 14中所有的元编程，只需要从fit库中使用fit::条件来编写:

template<class T>
std::string optionalToString(T* x)
{
    return fit::conditional(
        [](auto* obj) -> decltype(obj->toString()) { return obj->toString(); },
        [](auto*) { return "toString not defined"; }
    )(x);
}

你也可以直接从lambdas中创建函数:

FIT_STATIC_LAMBDA_FUNCTION(optionalToString) = fit::conditional(
    [](auto* obj) -> decltype(obj->toString(), std::string()) { return obj->toString(); },
    [](auto*) -> std::string { return "toString not defined"; }
);

然而，如果你使用的编译器不支持泛型lambdas，你将不得不编写单独的函数对象:

struct withToString
{
    template<class T>
    auto operator()(T* obj) const -> decltype(obj->toString(), std::string())
    {
        return obj->toString();
    }
};

struct withoutToString
{
    template<class T>
    std::string operator()(T*) const
    {
        return "toString not defined";
    }
};

FIT_STATIC_FUNCTION(optionalToString) = fit::conditional(
    withToString(),
    withoutToString()
);