3

我想获取参数包中最后一个可能类的成员变量的值。

例如。我想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;
}

但是，当我尝试编译此代码时，出现错误

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;

如果我注释我从 main 调用 getLastB 的行，代码编译并且它确实给出了 4 个打印语句的预期值。

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

有什么想法我可能在这里做错了吗？

4

2 回答 2

3

您需要从中删除参考T

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

b如果最后一个包含或b不是类型，您的代码也会失败int。

如果你可以使用 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;
}
于 2021-01-18T16:16:18.583 回答
2

因为我不确定c++2a我是否会将其作为单独的答案。

有了Concept支持，您可以更轻松地编写它（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;
    }
}
于 2021-01-19T13:15:15.417 回答