c++ - 当 static_cast'ing 为仅移动类型时，Clang 与 GCC

Question

考虑以下简单的仅移动类:

struct bar {
    constexpr bar() = default;
    bar(bar const&) = delete;
    bar(bar&&)      = default;
    bar& operator=(bar const&) = delete;
    bar& operator=(bar&&)      = default;
};

现在，让我们创建一个包装器:

template <class T>
struct box {
    constexpr box(T&& x)
        : _payload{std::move(x)}
    {}

    constexpr explicit operator T() &&
    {
        return std::move(_payload);
    }

  private:
    T _payload;
};

还有一个测试:

int main()
{
    auto x = box<bar>{bar{}};
    auto y = static_cast<bar&&>(std::move(x));
}

如果使用 -std=c++14 或更高版本编译，Clang-6.0 对这段代码很满意。然而，GCC-8.1 会产生以下错误:

<source>: In function 'int main()':
<source>:29:45: error: invalid static_cast from type 'std::remove_reference<box<bar>&>::type' {aka 'box<bar>'} to type 'bar&&'
     auto y = static_cast<bar&&>(std::move(x));

这是一个 link to compiler explorer尝试一下。

所以我的问题是，谁对谁错？

Answer 1

简化示例:

struct bar 
{
    bar() = default;
    bar(bar const&) = delete;
    bar(bar&&) = default;
};

struct box 
{
    explicit operator bar() && { return bar{}; }
};

int main() { static_cast<bar&&>(box{}); }

live on godbolt.org

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首先，让我们看看转换运算符是显式意味着什么:

A conversion function may be explicit, in which case it is only considered as a user-defined conversion for direct-initialization. Otherwise, user-defined conversions are not restricted to use in assignments and initializations.

static_cast 是否被视为直接初始化？

The initialization that occurs in the forms

T x(a);
T x{a};

as well as in new expressions, static_­cast expressions, functional notation type conversions, mem-initializers, and the braced-init-list form of a condition is called direct-initialization.

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由于 static_cast 是直接初始化，转换运算符是否标记为 explicit 并不重要。但是，删除 explicit 会使代码在 g++ 和 clang++ 上编译:live example on godbolt.org .

这让我相信这是一个 g++ 错误，因为 explicit 在这里有所作为......当它不应该的时候。