从参数包的最后一个可能的类中获取成员

Question

我想获得参数包中最后一个可能的类的成员变量的值。

例如。我希望getLastB(a_, b_, c_, d_)在下面的代码中返回100 (c_b的值)，而不是40 (b_.b的值)

#include <type_traits>
#include <iostream>
#define ADD_HAS_MEM_VAR_CHECKER(var, name)                              
template<typename T>                                                
struct name {                                                       
typedef char yes[1];                                            
typedef char no [2];                                            
template <typename _1> static yes &chk(decltype(_1::b));        
template <typename   > static no  &chk(...);                    
static bool const value = sizeof(chk<T>(0)) == sizeof(yes);     
}

ADD_HAS_MEM_VAR_CHECKER(b, has_b);
template<typename Arg, typename... Args>
struct has_b_parampack {
static bool const value = has_b<Arg>::value or has_b_parampack<Args...>::value;
};
template<typename Arg>
struct has_b_parampack<Arg> {
static bool const value = has_b<Arg>::value;
};

template <typename Arg, typename... Args>
typename std::enable_if<has_b_parampack<Args...>::value, int>::type
getLastB(Arg first, Args&&... args) {
return getLastB(args...);
}
template <typename Arg, typename... Args>
typename std::enable_if<!has_b_parampack<Args...>::value, int>::type
getLastB(Arg first, Args&&... args) {
return first.b;
}

struct A {int a = 10;};
struct B {int b = 40;};
struct C {int c = 30;int b = 100;};
struct D {int d = 30;};

int main() {
A a_;
B b_;
C c_;
D d_;
std::cout << "has B parampack A, B, C, D= " << has_b_parampack<A, B, C, D>::value << std::endl;
std::cout << "has B parampack B, C, D= " << has_b_parampack<B, C, D>::value << std::endl;
std::cout << "has B parampack C, D= " << has_b_parampack<C, D>::value << std::endl;
std::cout << "has B parampack D= " << has_b_parampack<D>::value << std::endl;
std::cout << "Last B = " << getLastB(a_, b_, c_, d_) << "n";
return 0;
}

然而，当我试图编译这段代码时，我得到一个错误

lastMemberOfParamPack.cpp: In instantiation of ‘typename std::enable_if<(! has_b_parampack<Args ...>::value), int>::type getLastB(Arg, Args&& ...) [with Arg = A; Args = {B&, C&, D&}; typename std::enable_if<(! has_b_parampack<Args ...>::value), int>::type = int]’:
lastMemberOfParamPack.cpp:58:56:   required from here
lastMemberOfParamPack.cpp:37:18: error: ‘struct A’ has no member named ‘b’
return first.b;

如果我注释了从main调用getLastB的行，代码将编译并给出4个print语句的预期值。

has B parampack A, B, C, D= 1
has B parampack B, C, D= 1
has B parampack C, D= 1
has B parampack D= 0

你知道我在这里可能做错了什么吗?

Answer 1

您需要从T中删除引用

static bool const value = sizeof(chk<typename std::remove_reference<T>::type>(0)) == sizeof(yes);

当最后一个包含b或b不是int类型时，您的代码也会失败。

---

如果你能使用c++ 17，你可以这样写:

template<typename T>                                               
struct has_b {                                                     
typedef char yes[1];
typedef char no [2];
template <typename S> static yes& chk(decltype(S::b));
template <typename  > static no & chk(...);
static bool const value = sizeof(chk<typename std::remove_reference<T>::type>(0)) == sizeof(yes);
};
template <typename Arg, typename... Args>
auto getLastB(Arg& first, Args&... args) {
if constexpr((has_b<Args>::value or...))
return getLastB(args...);
else
return first.b;
}

Answer 2

因为我不太确定c++2a，我会把这个作为单独的答案。

---

有了Concept的支持，你可以更容易地写(主要是HasB部分)

template <typename T>
concept HasB = requires(T t){t.b;};
template <typename T, typename... Ts> 
requires (HasB<T> or ... or HasB<Ts>) // optional SFINAE
auto getLastB(T& first, Ts&... rest){
if constexpr((HasB<Ts> or...))
return getLastB(rest...);
else if constexpr(HasB<T>) // optional check
return first.b; 
else{ // optional else
static_assert(HasB<T>,"no");
return 1;
}
}