是否有可能编写一个模板，根据某个成员函数是否定义在类上而改变行为?
下面是我想写的一个简单的例子:
template<class T>
std::string optionalToString(T* obj)
{
if (FUNCTION_EXISTS(T->toString))
return obj->toString();
else
return "toString not defined";
}
因此，如果类T定义了toString()，那么它就使用它;否则，它就不会。我不知道如何做的神奇部分是“FUNCTION_EXISTS”部分。
当前回答
这个解决方案怎么样?
#include <type_traits>
template <typename U, typename = void> struct hasToString : std::false_type { };
template <typename U>
struct hasToString<U,
typename std::enable_if<bool(sizeof(&U::toString))>::type
> : std::true_type { };
其他回答
下面是我的版本，它可以任意处理所有可能的成员函数重载，包括模板成员函数，可能带有默认实参。当成员函数调用某个类类型时，它区分了3种互斥的情况，给定的arg类型:(1)有效，或(2)模糊，或(3)不可用。使用示例:
#include <string>
#include <vector>
HAS_MEM(bar)
HAS_MEM_FUN_CALL(bar)
struct test
{
void bar(int);
void bar(double);
void bar(int,double);
template < typename T >
typename std::enable_if< not std::is_integral<T>::value >::type
bar(const T&, int=0){}
template < typename T >
typename std::enable_if< std::is_integral<T>::value >::type
bar(const std::vector<T>&, T*){}
template < typename T >
int bar(const std::string&, int){}
};
现在你可以这样使用它:
int main(int argc, const char * argv[])
{
static_assert( has_mem_bar<test>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(char const*,long)>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(std::string&,long)>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(std::vector<int>, int*)>::value , "");
static_assert( has_no_viable_mem_fun_call_bar<test(std::vector<double>, double*)>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(int)>::value , "");
static_assert( std::is_same<void,result_of_mem_fun_call_bar<test(int)>::type>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(int,double)>::value , "");
static_assert( not has_valid_mem_fun_call_bar<test(int,double,int)>::value , "");
static_assert( not has_ambiguous_mem_fun_call_bar<test(double)>::value , "");
static_assert( has_ambiguous_mem_fun_call_bar<test(unsigned)>::value , "");
static_assert( has_viable_mem_fun_call_bar<test(unsigned)>::value , "");
static_assert( has_viable_mem_fun_call_bar<test(int)>::value , "");
static_assert( has_no_viable_mem_fun_call_bar<test(void)>::value , "");
return 0;
}
下面是用c++11编写的代码，但是，你可以很容易地将它移植到具有typeof扩展的非c++11(例如gcc)。你可以用你自己的宏替换HAS_MEM宏。
#pragma once
#if __cplusplus >= 201103
#include <utility>
#include <type_traits>
#define HAS_MEM(mem) \
\
template < typename T > \
struct has_mem_##mem \
{ \
struct yes {}; \
struct no {}; \
\
struct ambiguate_seed { char mem; }; \
template < typename U > struct ambiguate : U, ambiguate_seed {}; \
\
template < typename U, typename = decltype(&U::mem) > static constexpr no test(int); \
template < typename > static constexpr yes test(...); \
\
static bool constexpr value = std::is_same<decltype(test< ambiguate<T> >(0)),yes>::value ; \
typedef std::integral_constant<bool,value> type; \
};
#define HAS_MEM_FUN_CALL(memfun) \
\
template < typename Signature > \
struct has_valid_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct has_valid_mem_fun_call_##memfun< T(Args...) > \
{ \
struct yes {}; \
struct no {}; \
\
template < typename U, bool = has_mem_##memfun<U>::value > \
struct impl \
{ \
template < typename V, typename = decltype(std::declval<V>().memfun(std::declval<Args>()...)) > \
struct test_result { using type = yes; }; \
\
template < typename V > static constexpr typename test_result<V>::type test(int); \
template < typename > static constexpr no test(...); \
\
static constexpr bool value = std::is_same<decltype(test<U>(0)),yes>::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename U > \
struct impl<U,false> : std::false_type {}; \
\
static constexpr bool value = impl<T>::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename Signature > \
struct has_ambiguous_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct has_ambiguous_mem_fun_call_##memfun< T(Args...) > \
{ \
struct ambiguate_seed { void memfun(...); }; \
\
template < class U, bool = has_mem_##memfun<U>::value > \
struct ambiguate : U, ambiguate_seed \
{ \
using ambiguate_seed::memfun; \
using U::memfun; \
}; \
\
template < class U > \
struct ambiguate<U,false> : ambiguate_seed {}; \
\
static constexpr bool value = not has_valid_mem_fun_call_##memfun< ambiguate<T>(Args...) >::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename Signature > \
struct has_viable_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct has_viable_mem_fun_call_##memfun< T(Args...) > \
{ \
static constexpr bool value = has_valid_mem_fun_call_##memfun<T(Args...)>::value \
or has_ambiguous_mem_fun_call_##memfun<T(Args...)>::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename Signature > \
struct has_no_viable_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct has_no_viable_mem_fun_call_##memfun < T(Args...) > \
{ \
static constexpr bool value = not has_viable_mem_fun_call_##memfun<T(Args...)>::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename Signature > \
struct result_of_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct result_of_mem_fun_call_##memfun< T(Args...) > \
{ \
using type = decltype(std::declval<T>().memfun(std::declval<Args>()...)); \
};
#endif
我也遇到过类似的问题:
一个模板类，可以从少数基类派生，其中一些基类具有某个成员，而另一些基类没有。
我解决它类似于“typeof”(Nicola Bonelli)的答案，但使用decltype，所以它在MSVS上编译和正确运行:
#include <iostream>
#include <string>
struct Generic {};
struct HasMember
{
HasMember() : _a(1) {};
int _a;
};
// SFINAE test
template <typename T>
class S : public T
{
public:
std::string foo (std::string b)
{
return foo2<T>(b,0);
}
protected:
template <typename T> std::string foo2 (std::string b, decltype (T::_a))
{
return b + std::to_string(T::_a);
}
template <typename T> std::string foo2 (std::string b, ...)
{
return b + "No";
}
};
int main(int argc, char *argv[])
{
S<HasMember> d1;
S<Generic> d2;
std::cout << d1.foo("HasMember: ") << std::endl;
std::cout << d2.foo("Generic: ") << std::endl;
return 0;
}
这个解决方案怎么样?
#include <type_traits>
template <typename U, typename = void> struct hasToString : std::false_type { };
template <typename U>
struct hasToString<U,
typename std::enable_if<bool(sizeof(&U::toString))>::type
> : std::true_type { };
我修改了https://stackoverflow.com/a/264088/2712152中提供的解决方案，使其更加通用。此外，由于它不使用任何新的c++ 11特性，我们可以将它与旧的编译器一起使用，并且应该也可以与msvc一起使用。但是编译器应该允许C99使用这个，因为它使用可变宏。
下面的宏可用于检查特定类是否具有特定类型定义。
/**
* @class : HAS_TYPEDEF
* @brief : This macro will be used to check if a class has a particular
* typedef or not.
* @param typedef_name : Name of Typedef
* @param name : Name of struct which is going to be run the test for
* the given particular typedef specified in typedef_name
*/
#define HAS_TYPEDEF(typedef_name, name) \
template <typename T> \
struct name { \
typedef char yes[1]; \
typedef char no[2]; \
template <typename U> \
struct type_check; \
template <typename _1> \
static yes& chk(type_check<typename _1::typedef_name>*); \
template <typename> \
static no& chk(...); \
static bool const value = sizeof(chk<T>(0)) == sizeof(yes); \
}
下面的宏可以用来检查一个特定的类是否有一个特定的成员函数，是否有给定数量的参数。
/**
* @class : HAS_MEM_FUNC
* @brief : This macro will be used to check if a class has a particular
* member function implemented in the public section or not.
* @param func : Name of Member Function
* @param name : Name of struct which is going to be run the test for
* the given particular member function name specified in func
* @param return_type: Return type of the member function
* @param ellipsis(...) : Since this is macro should provide test case for every
* possible member function we use variadic macros to cover all possibilities
*/
#define HAS_MEM_FUNC(func, name, return_type, ...) \
template <typename T> \
struct name { \
typedef return_type (T::*Sign)(__VA_ARGS__); \
typedef char yes[1]; \
typedef char no[2]; \
template <typename U, U> \
struct type_check; \
template <typename _1> \
static yes& chk(type_check<Sign, &_1::func>*); \
template <typename> \
static no& chk(...); \
static bool const value = sizeof(chk<T>(0)) == sizeof(yes); \
}
我们可以使用上面的两个宏来检查has_typedef和has_mem_func:
class A {
public:
typedef int check;
void check_function() {}
};
class B {
public:
void hello(int a, double b) {}
void hello() {}
};
HAS_MEM_FUNC(check_function, has_check_function, void, void);
HAS_MEM_FUNC(hello, hello_check, void, int, double);
HAS_MEM_FUNC(hello, hello_void_check, void, void);
HAS_TYPEDEF(check, has_typedef_check);
int main() {
std::cout << "Check Function A:" << has_check_function<A>::value << std::endl;
std::cout << "Check Function B:" << has_check_function<B>::value << std::endl;
std::cout << "Hello Function A:" << hello_check<A>::value << std::endl;
std::cout << "Hello Function B:" << hello_check<B>::value << std::endl;
std::cout << "Hello void Function A:" << hello_void_check<A>::value << std::endl;
std::cout << "Hello void Function B:" << hello_void_check<B>::value << std::endl;
std::cout << "Check Typedef A:" << has_typedef_check<A>::value << std::endl;
std::cout << "Check Typedef B:" << has_typedef_check<B>::value << std::endl;
}
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;
}
