Question

花了1或2个小时来隔离由元编程混乱所包围的编译错误，生成可怕的编译消息，这是一个简单的例子来说明我的问题：

#include <iostream>
#include <type_traits>
#include <array>
#include <utility>
#include <tuple>

template <class Crtp, class... Types>
struct Base
{
    Base(const Types&... rhs) : 
        data(std::forward_as_tuple(rhs...)) {;}
    std::tuple<Types...> data;
};

struct Derived 
: public Base<Derived, std::array<double, 3>>
{
    template <class... Args> 
    Derived(Args&&... args) :
        Base<Derived, std::array<double, 3>>(std::forward<Args>(args)...) {;}
};

int main(int argc, char* argv[])
{
    Derived a(std::array<double, 3>({{1, 2, 3}})); 
    Derived b(a);
    Derived c(std::array<double, 3>()); 
    Derived d(c); // Not working : why ?
    return 0;
}

这是用g ++ 4.8.1编译的，当我尝试复制c中的d而不是a中的b时，我不明白为什么编译器会抱怨

这是错误：

main.cpp: In instantiation of ‘Derived::Derived(Args&& ...) [with Args = {Derived (&)(std::array<double, 3ul> (*)())}]’:
main.cpp:28:16:   required from here
main.cpp:20:73: error: no matching function for call to ‘Base<Derived, std::array<double, 3ul> >::Base(Derived (&)(std::array<double, 3ul> (*)()))’
         Base<Derived, std::array<double, 3>>(std::forward<Args>(args)...) {;}
                                                                         ^
main.cpp:20:73: note: candidates are:
main.cpp:10:5: note: Base<Crtp, Types>::Base(const Types& ...) [with Crtp = Derived; Types = {std::array<double, 3ul>}]
     Base(const Types&... rhs) : 
     ^
main.cpp:10:5: note:   no known conversion for argument 1 from ‘Derived(std::array<double, 3ul> (*)())’ to ‘const std::array<double, 3ul>&’
main.cpp:8:8: note: constexpr Base<Derived, std::array<double, 3ul> >::Base(const Base<Derived, std::array<double, 3ul> >&)
 struct Base
        ^
main.cpp:8:8: note:   no known conversion for argument 1 from ‘Derived(std::array<double, 3ul> (*)())’ to ‘const Base<Derived, std::array<double, 3ul> >&’
main.cpp:8:8: note: constexpr Base<Derived, std::array<double, 3ul> >::Base(Base<Derived, std::array<double, 3ul> >&&)
main.cpp:8:8: note:   no known conversion for argument 1 from ‘Derived(std::array<double, 3ul> (*)())’ to ‘Base<Derived, std::array<double, 3ul> >&&’

Answer 1

这是most vexing parse：

Derived c(std::array<double, 3>());

是一个函数c的声明，它返回一个Derived并接受一个未命名的类型为的参数指向函数，该函数不带参数并返回std::array<double, 3> 。因此Derived d(c)尝试从函数Derived调用c构造函数。这就是海湾合作委员会在这里所说的：

main.cpp: In instantiation of ‘Derived::Derived(Args&& ...) [with Args = {Derived (&)(std::array<double, 3ul> (*)())}]’:

试试这个：

Derived c{std::array<double, 3>{}};

Answer 2

适用于：

Derived c(std::array<double, 3> {});

编译器会考虑

中的参数

Derived c(std::array<double, 3>());

是一个功能。

clang对此发出警告：
!!warning: parentheses were disambiguated as a function declarator.

Answer 3

首先使Derived构造函数显式化。如果这没有帮助添加复制构造函数，但我不认为这是必需的

无法复制从默认构造的数组构造的对象

3 个答案: