c++ - std::variant 转换构造函数行为

Question

C++17 std::variant<class... Types>有一个转换构造函数

template< class T >
constexpr variant(T&& t) noexcept(/* see below */);

(http://en.cppreference.com/w/cpp/utility/variant/variant 中的第 4 个)。它的描述是一堵相当坚不可摧的文字墙。这是否意味着变体有一堆

template< class T_i > constexpr variant(T_i&& t) noexcept;

构造器，类型中的每个 T_i 一个？

Answer 1

让我们分解一下描述:

Constructs a variant holding the alternative type T<sub>j</sub> that would be selected by overload resolution for the expression F(std::forward<T>(t)) if there was an overload of imaginary function F(T<sub>i</sub>) for every T<sub>i</sub> from Types... in scope at the same time.

假设我们在文档中有示例:

variant<string, bool> x("abc"); // OK, but chooses bool

• Types...是<string, bool>

---

if there was an overload of imaginary function F(T<sub>i</sub>) for every T<sub>i</sub> from Types...

int F(string) { return 0; }
int F(bool)   { return 1; }

---

selected by overload resolution for the expression F(std::forward<T>(t))

template <typename T>
void select(T&& t)
{
    std::cout << F(std::forward<T>(t)) << '\n';
}

int main()
{
    select("abc"); // prints `1`
}

live example on wandbox

---

Constructs a variant holding the alternative type T<sub>j</sub> [...]

选择的替代类型 T<sub>j</sub>因此是bool .