std::chrono::duration::duration() 怎么可能是 constexpr？

Question

std::chrono::duration的默认构造函数定义如下：

constexpr duration() = default;

（例如，参见 cppreference.com 或 libstdc++ 源代码。）

但是，cppreference.com also says this 关于 constexpr 个构造函数：

A constexpr constructor must satisfy the following requirements:

...

every base class and every non-static member must be initialized, either in the constructors initialization list or by a member brace-or-equal initializer. In addition, every constructor involved must be a constexpr constructor and every clause of every brace-or-equal initializer must be a constant expression

如果我对默认构造函数感到困惑，cppreference.com seems to say = default 带来的默认构造函数与隐式默认构造函数的定义没有区别。

然而，（大多数）持续时间的 rep 类型是裸整数类型。因此， duration 的显式 = default 默认构造函数不应该等同于

constexpr duration() {}

哪一个当然会让 duration::rep 类型的整型成员变量未初始化？而且，事实上，duration 的标准行为不就是默认构造的值是未初始化的吗？（但我找不到明确说明这一点的参考资料。）

那么，如果 duration 的 = default 构造函数让非静态成员变量未初始化，那么它怎么会是 constexpr 呢？我错过了什么？

Answer 1

7.1.5 constexpr 说明符 [dcl.constexpr] 表示：

The definition of a constexpr constructor shall satisfy the following requirements:

the class shall not have any virtual base classes;

for a defaulted copy/move constructor, the class shall not have a mutable subobject that is a variant member;

each of the parameter types shall be a literal type;

its function-body shall not be a function-try-block;

In addition, either its function-body shall be = delete, or it shall satisfy the following requirements:

either its function-body shall be = default, or the compound-statement of its function-body shall satisfy the requirements for a function-body of a constexpr function;

every non-variant non-static data member and base class sub-object shall be initialized (12.6.2);

if the class is a union having variant members (9.5), exactly one of them shall be initialized;

if the class is a union-like class, but is not a union, for each of its anonymous union members having variant members, exactly one of them shall be initialized;

for a non-delegating constructor, every constructor selected to initialize non-static data members and base class sub-objects shall be a constexpr constructor;

for a delegating constructor, the target constructor shall be a constexpr constructor.

简而言之，只要满足上述其他要求，= default 就是 constexpr 默认构造函数的有效定义。

那么这如何与 未初始化的 结构一起工作？

没有。

例如：

constexpr seconds x1{};

上面的工作并将 x1 初始化为 0s。然而：

constexpr seconds x2;

error: default initialization of an object of const type 'const seconds'
       (aka 'const duration<long long>') without a user-provided default
        constructor
    constexpr seconds x2;
                      ^
                        {}
1 error generated.

因此，要创建一个 constexpr 默认构造的 duration，您必须 零初始化 它。 = default 实现允许使用 {}.

进行零初始化

完成工作演示：

template <class Rep>
class my_duration
{
    Rep rep_;
public:
    constexpr my_duration() = default;
};


int
main()
{
    constexpr my_duration<int> x{};
}

有趣的边栏

我在写这个答案的过程中学到了一些东西，想分享一下：

我一直想知道为什么以下方法不起作用：

using Rep = int;

class my_duration
{
    Rep rep_;
public:
    constexpr my_duration() = default;
};


int
main()
{
    constexpr my_duration x{};
}

error: defaulted definition of default constructor is not constexpr
        constexpr my_duration() = default;
        ^

为什么将此 class 设为非模板会破坏 constexpr 默认构造函数？！

(更新： 现在用 C++20 编译)

然后我试了这个：

using Rep = int;

class my_duration
{
    Rep rep_;
public:
    my_duration() = default;  // removed constexpr
};


int
main()
{
    constexpr my_duration x{};
}

编译器再次喜欢它。

~~如果还没有关于此的 CWG 问题，可能应该有。行为似乎有点不一致。这可能只是因为我们（整个行业）仍在学习 constexpr.~~ Fixed in C++20.

std::chrono::duration::duration() 怎么可能是 constexpr？

How can std::chrono::duration::duration() be constexpr?

c++

c++11

chrono