这是我在c++ 20中发现的最简洁的方法，非常接近你的问题:

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

在godbolt上观看:https://gcc.godbolt.org/z/5jb1d93Ms

下面是工作代码的示例。

template<typename T>
using toStringFn = decltype(std::declval<const T>().toString());

template <class T, toStringFn<T>* = nullptr>
std::string optionalToString(const T* obj, int)
{
    return obj->toString();
}

template <class T>
std::string optionalToString(const T* obj, long)
{
    return "toString not defined";
}

int main()
{
    A* a;
    B* b;

    std::cout << optionalToString(a, 0) << std::endl; // This is A
    std::cout << optionalToString(b, 0) << std::endl; // toString not defined
}

toStringFn<T>* = nullptr将启用带有额外int参数的函数，该函数的优先级高于使用0调用时需要很长时间的函数。

你可以对函数使用相同的原则，如果函数被实现，返回true。

template <typename T>
constexpr bool toStringExists(long)
{
    return false;
}

template <typename T, toStringFn<T>* = nullptr>
constexpr bool toStringExists(int)
{
    return true;
}


int main()
{
    A* a;
    B* b;

    std::cout << toStringExists<A>(0) << std::endl; // true
    std::cout << toStringExists<B>(0) << std::endl; // false
}

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

template<class T>
auto optionalToString(T* obj)
->decltype( std::string() )
{
     throw "Error!";
}

pre -c++20，简单用例的简单选项:

如果你知道你的类是默认可构造的，我们可以使语法更简单。

我们将从最简单的情况开始:默认可构造对象，并且我们知道预期的返回类型。实例方法:

int foo ();

我们可以写出没有declval的类型trait:

template <auto v>
struct tag_v
{
    constexpr static auto value = v;
};

template <class, class = int>
struct has_foo_method : tag_v<false> {};

template <class T>
struct has_foo_method <T, decltype(T().foo())>
    : tag_v<true> {};

demo

注意，我们将默认类型设置为int，因为这是foo的返回类型。

如果有多个可接受的返回类型，则向decltype添加第二个参数，该参数与默认类型相同，覆盖第一个参数:

decltype(T().foo(), int())

demo

(这里的int类型不重要-我使用它是因为它只有3个字母)

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;
}