【发布时间】:2017-06-15 15:51:39
【问题描述】:
尝试创建一种方法来识别给定类是否具有可以调用的给定函数,并返回某种类型。
知道我在这里做错了什么吗?有没有更好的方法来确定给定方法是否可以在给定类的情况下调用?
#include <string>
#include <type_traits>
#define GENERATE_HAS_MEMBER_FUNC(func, rettype) \
template<typename T, class Enable = void> struct has_##func; \
template<typename T, class U> struct has_##func : std::false_type {}; \
template<typename T> \
struct has_##func<T, \
typename std::enable_if<std::is_same< \
typename std::result_of<decltype (&T::func)(T)>::type, \
rettype>::value>::type> : std::true_type{}; \
template<class T> constexpr bool has_##func##_v = has_##func<T>::value;
GENERATE_HAS_MEMBER_FUNC(str, std::string)
GENERATE_HAS_MEMBER_FUNC(str2, std::string)
GENERATE_HAS_MEMBER_FUNC(funca, std::string)
GENERATE_HAS_MEMBER_FUNC(strK, std::string)
GENERATE_HAS_MEMBER_FUNC(fancy, std::string)
GENERATE_HAS_MEMBER_FUNC(really, std::string)
struct A1 {
virtual std::string str() const { return ""; }
std::string strK() const { return ""; }
virtual std::string fancy()=0;
};
struct A2 : A1 {
std::string str() const override { return ""; }
std::string funca();
std::string fancy() override { return ""; }
std::string really(int a=0) const { return std::to_string(a); }
};
int main() {
static_assert(has_str_v<A1>,
"A1::str is virtual method with impl on base"); // MSVC: NO, clang: OK, GCC: NO
static_assert(has_strK_v<A1>,
"A1::strK is implemented inline "); // MSVC: NO, clang: OK, GCC: NO
static_assert(has_fancy_v<A1>,
"A1::fancy is a pure virtual method on base"); // MSVC: NO, clang: OK, GCC: NO
static_assert(!has_really_v<A1>,
"A1::really doesn't exist in A1"); // MSVC: OK, clang: OK, GCC: OK
static_assert(has_str_v<A2>,
"A2::str is override method "); // MSVC: OK, clang: OK, GCC: OK
static_assert(!has_str2_v<A2>,
"A2::str2 does not exist in A2"); // MSVC: NO, clang: OK, GCC: OK
static_assert(has_funca_v<A2>,
"A2::funca is defined (no impl) in A2"); // MSVC: OK, clang: OK, GCC: OK
static_assert(has_strK_v<A2>,
"A2::strK is implemented method on base"); // MSVC: OK, clang: OK, GCC: OK
static_assert(has_fancy_v<A2>,
"A1::fancy is a override of pure virtual method of base"); // MSVC: OK, clang: OK, GCC: OK
static_assert(has_really_v<A2>,
"A2::really has default param (can be invoked without params)"); // MSVC: OK, clang: NO, GCC: NO
return 0;
}
这个实现的一些惊喜。
编辑: 在尝试实施@Jarod42 和@Vittorio Romeo 时,很棒的建议:
#define GENERATE_HAS_MEMBER_FUNC(func, rettype) \
template<class T> using _has_##func_chk = \
decltype(std::declval<T &>().func()); \
template<class T> constexpr bool has_##func##_v = \
is_detected_exact_v<rettype, _has_##func_chk, T>;
现在两个测试用例在 VS2015 上仍然失败(没有任何意义): static_assert(!has_really_v, "A1::really 不存在于 A1"); static_assert(!has_str2_v, "A2::str2 不存在于 A2");
我可能遗漏了一些愚蠢的东西......有什么线索吗?
【问题讨论】:
-
你看过
std::is_detected(C++17,但可以在C++11中实现)吗?
标签: c++ c++14 typetraits