[英]How to get a reference to variant’s value?
I have std::variant
where all classes are derived from the same base. 我有std::variant
,其中所有类均源自同一基数。 I want to cast variant to base. 我想将变体转换为基础。
return std::visit( []( const Base& b ) { return b; }, v );
This compiles but gives warning C4172: returning address of local variable or temporary 这会编译,但会发出警告C4172:局部变量或临时变量的返回地址
Is there a way to visit std::variant
in place, without making local or temporary copies? 有没有一种方法可以访问std::variant
,而不进行本地或临时副本?
Or if it's impossible, how can I cast the value to void*
so I can use static_cast
? 或者,如果不可能,如何将值static_cast
为void*
以便可以使用static_cast
?
Update: I thought the example should be obvious, but it's not, here's the complete repro: 更新:我认为该示例应该很明显,但是事实并非如此,这是完整的副本:
#include <variant>
struct Base {};
struct A : Base {};
struct B : Base {};
const Base& cast( const std::variant<A, B>& v )
{
return std::visit( []( Base const& b ) { return b; }, v );
}
int main()
{
std::variant<A, B> v{ A{} };
const auto& b = cast( v );
}
Lambdas have return type deduction, but they deduce the return type by value. Lambda具有返回类型推导,但是它们通过值推导返回类型。 It's as if they are a function returning auto
, not decltype(auto)
. 好像它们是一个返回auto
而不是decltype(auto)
的函数。 If you want to return by reference, you need to specify the return type. 如果要通过引用返回,则需要指定返回类型。
Thus, [](const Base& b) { return b; }
因此, [](const Base& b) { return b; }
[](const Base& b) { return b; }
returns by value, copying b
. [](const Base& b) { return b; }
按值返回,复制b
。 Explicitly specify the return type to force it to return by reference: 明确指定返回类型,以强制其通过引用返回:
const Base& cast( const std::variant<A, B>& v )
{
return std::visit( []( Base const& b ) -> Base const& { return b; }, v );
}
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