问题描述
我无法在Ubuntu 20.04.1 / gcc 9.3.0上编译以下代码。 根据{{3}},它可以使用gcc 10.x进行编译, 和gcc 7.x,但给出:
In file included from /usr/include/c++/9/variant:36,from test.cpp:1:
/usr/include/c++/9/type_traits: In instantiation of ‘struct std::__is_nt_default_constructible_atom<Foo>’:
/usr/include/c++/9/type_traits:945:12: required from ‘struct std::__is_nt_default_constructible_impl<Foo,false>’
/usr/include/c++/9/type_traits:131:12: required from ‘struct std::__and_<std::is_default_constructible<Foo>,std::__is_nt_default_constructible_impl<Foo,false> >’
/usr/include/c++/9/type_traits:951:12: required from ‘struct std::is_nothrow_default_constructible<Foo>’
/usr/include/c++/9/type_traits:2965:25: required from ‘constexpr const bool std::is_nothrow_default_constructible_v<Foo>’
/usr/include/c++/9/variant:301:4: [ skipping 2 instantiation contexts,use -ftemplate-backtrace-limit=0 to disable ]
/usr/include/c++/9/type_traits:883:12: recursively required from ‘constexpr std::variant<_Types>::variant() [with _Types = {Foo,Boo}]’
/usr/include/c++/9/type_traits:883:12: required from ‘struct std::is_constructible<std::variant<Foo,Boo> >’
/usr/include/c++/9/type_traits:889:12: required from ‘struct std::is_default_constructible<std::variant<Foo,Boo> >’
/usr/include/c++/9/type_traits:2921:25: required from ‘constexpr const bool std::is_default_constructible_v<std::variant<Foo,Boo> >’
/usr/include/c++/9/variant:273:4: required from ‘constexpr const bool std::__detail::__variant::_Traits<std::variant<Foo,Boo>,std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<c\
har> > >::_S_default_ctor’
/usr/include/c++/9/variant:1219:11: required from ‘class std::variant<std::variant<Foo,std::allocator<char> > >’
test.cpp:32:53: required from here
/usr/include/c++/9/type_traits:931:47: error: ‘Foo::Foo()’ is private within this context
931 | : public integral_constant<bool,noexcept(_Tp())>
| ^~~~~
test.cpp:10:3: note: declared private here
10 | Foo() noexcept {}
| ^~~
使用gcc 8.x和9.x。波纹管有什么问题吗?
请注意,如果在下面的代码中删除template <bool X>
,
所有编译都很好。或者,我可以删除main中的代码,并且所有编译都可以。
#include <variant>
#include <string>
struct Foo {
public:
explicit Foo(int) noexcept {}
Foo(Foo &&) noexcept = default;
Foo &operator=(Foo &&) = default;
private:
Foo() noexcept {}
};
struct Boo {
public:
explicit Boo(int) noexcept {}
Boo(Boo &&) noexcept = default;
Boo &operator=(Boo &&) = default;
private:
Boo() noexcept {}
};
template<bool X>
std::variant<Foo,Boo> g(int v,int x) {
return v == 0 ? std::variant<Foo,Boo>{Foo{x}} :
std::variant<Foo,Boo>{Boo{x}};
}
int main()
{
std::variant<std::variant<Foo,std::string> err{std::string("aaa")};
}
解决方法
根据标准,在这种情况下没有DefaultConstructible要求。
允许变体本身是Non-DefaultConstructible的(内部和外部变体在此示例中都是这种情况)。
如果Foo()私有构造函数被删除或删除,则code works(按预期):
struct Foo {
public:
explicit Foo(int) noexcept {}
Foo(Foo &&) noexcept = default;
Foo &operator=(Foo &&) = default;
private:
// Foo() noexcept {} -> comment out works
Foo() = deleted; // -> works also.
};
因此,这显然是编译器或标准库中的实现错误。