在ostream‘<<’中使用std::endl --带有包含变量成员的可变模板类的操作符重载会导致编译器错误

Question

我有下面的代码，其中我定义了一个ostream <<-operator重载变量模板参数类型VariantContainer，它包含一个std::variant并使用来自VariantContainer for std::variant的可变模板参数

#include <iostream>
#include <variant>

template<typename... VARIANT_TYPES>
struct VariantContainer {
    std::variant<VARIANT_TYPES...> var;
};

template<typename... VARIANT_TYPES>
std::ostream& operator<<(std::ostream& os, const VariantContainer<VARIANT_TYPES...>& inst) {
    std::visit([&os](auto&& elem){os<<elem;}, inst.var);
    return os;
}

int main() {
    //If the following line is commented out, a compiler error is raised
    //std::cout << std::endl;
}

此示例编译在gcc-8、gcc-9和with 10上，其中--std=c++17 (https://godbolt.org/z/5qvT3f).但是，当我启用std::cout-line时，会得到以下编译器错误：

In file included from <source>:2:
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant: In instantiation of 'constexpr const bool std::__detail::__variant::_Traits<>::_S_default_ctor':
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:1279:11:   required from 'class std::variant<>'
<source>:6:36:   required from 'struct VariantContainer<>'
<source>:17:23:   required from here
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:286:4: error: invalid use of incomplete type 'struct std::__detail::__variant::_Nth_type<0>'
  286 |    is_default_constructible_v<typename _Nth_type<0, _Types...>::type>;
      |    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:61:12: note: declaration of 'struct std::__detail::__variant::_Nth_type<0>'
   61 |     struct _Nth_type;
      |            ^~~~~~~~~
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant: In instantiation of 'class std::variant<>':
<source>:6:36:   required from 'struct VariantContainer<>'
<source>:17:23:   required from here
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:1298:39: error: static assertion failed: variant must have at least one alternative
 1298 |       static_assert(sizeof...(_Types) > 0,
      |                     ~~~~~~~~~~~~~~~~~~^~~
In file included from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/bits/move.h:57,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/bits/nested_exception.h:40,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/exception:148,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/ios:39,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/ostream:38,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/iostream:39,
                 from <source>:1:
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/type_traits: In substitution of 'template<bool _Cond, class _Tp> using enable_if_t = typename std::enable_if::type [with bool _Cond = false; _Tp = void]':
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:1361:2:   required from 'class std::variant<>'
<source>:6:36:   required from 'struct VariantContainer<>'
<source>:17:23:   required from here
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/type_traits:2554:11: error: no type named 'type' in 'struct std::enable_if<false, void>'
 2554 |     using enable_if_t = typename enable_if<_Cond, _Tp>::type;
      |           ^~~~~~~~~~~
ASM generation compiler returned: 1
In file included from <source>:2:
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant: In instantiation of 'constexpr const bool std::__detail::__variant::_Traits<>::_S_default_ctor':
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:1279:11:   required from 'class std::variant<>'
<source>:6:36:   required from 'struct VariantContainer<>'
<source>:17:23:   required from here
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:286:4: error: invalid use of incomplete type 'struct std::__detail::__variant::_Nth_type<0>'
  286 |    is_default_constructible_v<typename _Nth_type<0, _Types...>::type>;
      |    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:61:12: note: declaration of 'struct std::__detail::__variant::_Nth_type<0>'
   61 |     struct _Nth_type;
      |            ^~~~~~~~~
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant: In instantiation of 'class std::variant<>':
<source>:6:36:   required from 'struct VariantContainer<>'
<source>:17:23:   required from here
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:1298:39: error: static assertion failed: variant must have at least one alternative
 1298 |       static_assert(sizeof...(_Types) > 0,
      |                     ~~~~~~~~~~~~~~~~~~^~~
In file included from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/bits/move.h:57,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/bits/nested_exception.h:40,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/exception:148,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/ios:39,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/ostream:38,
                 from /opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/iostream:39,
                 from <source>:1:
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/type_traits: In substitution of 'template<bool _Cond, class _Tp> using enable_if_t = typename std::enable_if::type [with bool _Cond = false; _Tp = void]':
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/variant:1361:2:   required from 'class std::variant<>'
<source>:6:36:   required from 'struct VariantContainer<>'
<source>:17:23:   required from here
/opt/compiler-explorer/gcc-10.2.0/include/c++/10.2.0/type_traits:2554:11: error: no type named 'type' in 'struct std::enable_if<false, void>'
 2554 |     using enable_if_t = typename enable_if<_Cond, _Tp>::type;
      |           ^~~~~~~~~~~
Execution build compiler returned: 1

在我看来，当编译器看到std::endl时，它试图用零模板参数实例化VariantContainer (这是不允许的，因为std::variant static_assert就是这样)。但是为什么编译器要这样做，我如何防止这种情况呢？

当然，我很清楚，我可以在to_string中创建一个VariantContainer方法来防止这种情况，或者提供一个更通用的std::ostream <<-operator版本(比如template<typename T> std::ostream& operator<<(std::ostream &, const T& ) {...}，并通过std::enable_if对其进行约束)。但我真的不喜欢这样。

Answer 1

std::endl是一个函数模板，重载也是如此。

使用

std::cout << std::endl;

我们必须采取(正确) 重载函数的地址。

为此，我们必须看到每个operator<<之间是否存在一些约束签名的因素。

确实有来自basic_stream的，特别是

basic_ostream& operator<<(
    std::basic_ios<CharT,Traits>& (*func)(std::basic_ios<CharT,Traits>&) );

(这是预期的最佳匹配)。

但我们还必须检查其他问题，特别是有问题的：

template<typename... VARIANT_TYPES>
std::ostream& operator<<(std::ostream& os, const VariantContainer<VARIANT_TYPES...>& inst);

而且，由于VARIANT_TYPES不能从函数集中还原，所以VARIANT_TYPES是空包。

现在检查VariantContainer<>(std::endl)是否会强制函数签名，因此实例化VariantContainer<>构造函数，因为std::variant<>导致了硬错误。

现在，可能的解决方案应该避免硬错误

template<typename... VARIANT_TYPES>
std::ostream& operator<<(std::ostream& os, const VariantContainer<VARIANT_TYPES...>& inst);

• 为类VariantContainer<>提供(有效或甚至不完全)专门化
• 将主模板更改为

template <typename T, typename... Ts> 
class VariantContainer;

(不需要更改operator <<，因为SFINAE将适用于无效的VariantContainer<>)

• 将operator<<更改为

template <typename T, typename... VARIANT_TYPES>
std::ostream& operator<<(std::ostream& os, const VariantContainer<T, VARIANT_TYPES...>& inst);

Answer 2

同时，我尝试像这样定义VariantContainer：

template<typename VARIANT_TYPE, typename... VARIANT_TYPES>
struct VariantContainer {
    std::variant<VARIANT_TYPE, VARIANT_TYPES...> var;
};

只是为了确保必须给出一个模板参数。

现在起作用了！

#include <iostream>
#include <variant>

struct A{
    int a;
};

struct B{
    int b;
};

std::ostream& operator<<(std::ostream& os, const A& inst) {os << inst.a; return os;}

std::ostream& operator<<(std::ostream& os, const B& inst) {os << inst.b; return os;}

template<typename VARIANT_TYPE, typename... VARIANT_TYPES>
struct VariantContainer {
    std::variant<VARIANT_TYPE, VARIANT_TYPES...> var;
};

template<typename... VARIANT_TYPES>
std::ostream& operator<<(std::ostream& os, const VariantContainer<VARIANT_TYPES...>& inst) {
    std::visit([&os](auto&& elem){os<<elem;}, inst.var);
    return os;
}

int main() {
    VariantContainer<A, B> insta{A{1}};
    VariantContainer<A, B> instb{B{1}};
    std::cout << "If I want to print something, I get a compiler error." << insta << " " << instb<< std::endl;
}