在花了 1 或 2 个小时来隔离编译错误,该编译错误被元编程困惑所包围,生成了可怕的编译消息,这里是一个最小的简单示例来说明我的问题:
#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,我不明白为什么当我尝试在 d
而不是 a< 中复制
在 c
时编译器会报错b
中。
这里是错误:
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> >&&’
最佳答案
这是 most vexing parse :
Derived c(std::array<double, 3>());
是一个函数的声明c
返回 Derived
并接受一个类型为 的未命名参数指向函数,该函数不接受任何参数并返回 std::array<double, 3>
。
因此Derived d(c)
尝试调用 Derived
来自函数的构造函数 c
.这就是 GCC 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>{}};
关于c++ - 无法复制从默认构造的数组构造的对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17486767/