默认情况下是否默认的构造函数/赋值为noexcept / constexpr？ [英] Is a defaulted constructor/assignment noexcept/constexpr by default?

问题描述

所以，我的问题很简单：

So, my question is simple:

将默认的类构造函数指定为 noexcept 有什么意义吗？或 constexpr （或您可能会想到的其他任何东西）？

Is there any point in specifying a defaulted class constructor as noexcept or constexpr (or any other thing you could thing of)?

struct foo
{
   foo() = default;
   // vs
   constexpr foo() noexcept = default;

   // same thing would apply for copy/move ctors and assignment operators
};

两者的行为是否相同？

Would the two behave the same way?

这是否取决于类是否为POD？
例如，在上面的示例中，两者的行为相同，而例如，如果我有一个私有成员 std :: vector< int> v = {1，2，3，4}; 使用类内赋值， foo（）=默认值； 默认不为 noexcept 而不是 constexpr 。

Does it depend on whether the class is POD? For example with the above example both would behave the same way, while if for example I had a private member std::vector<int> v = { 1, 2, 3, 4 }; which uses in-class assignment, foo() = default; would by default not be noexcept and not constexpr.

通过编写 foo（）=默认值； 编译器是否会选择最佳版本： noexcept （如果可能）和 constexpr ，如果可能，等等？

By writing foo() = default; does the compiler just pick the best version: noexcept if possible and constexpr if possible, etc?

推荐答案

[dcl.fct.def.default] / 2-3 ：

2未定义为删除的显式默认函数只有在隐式声明$ b $的情况下，才可以将
声明为 constexpr 。 b为 constexpr 。如果某个函数在其第一个
声明中显式默认为

2 An explicitly-defaulted function that is not defined as deleted may be declared constexpr only if it would have been implicitly declared as constexpr. If a function is explicitly defaulted on its first declaration,

• ，则该函数隐式视为 constexpr 如果隐式声明是，并且


• 它具有与隐式声明相同的异常规范（[except.spec]）。 li>
  

• it is implicitly considered to be constexpr if the implicit declaration would be, and,
• it has the same exception specification as if it had been implicitly declared ([except.spec]).

3如果使用
exception-specification 声明的默认值明确声明的函数不是与
兼容（[except.spec]）隐式声明的异常规范，然后

3 If a function that is explicitly defaulted is declared with an exception-specification that is not compatible ([except.spec]) with the exception specification of the implicit declaration, then

• 在其第一个声明中明确默认为默认值，它定义为已删除；

• if the function is explicitly defaulted on its first declaration, it is defined as deleted;

否则，程序格式不正确。

otherwise, the program is ill-formed.

换句话说， foo（）=默认值； ，它必须是 foo 的默认构造函数的第一个声明，如果可能的话，将是 constexpr 和 无例外。显式编写 constexpr 和 noexcept 仍然有用；它的意思是如果不能 constexpr / noexcept 大叫我。

In other words, foo() = default;, which is necessarily the first declaration of foo's default constructor, will be "constexpr if possible" and "noexcept if possible". Explicitly writing constexpr and noexcept is still useful; it means "yell at me if it can't be constexpr/noexcept".

这篇关于默认情况下是否默认的构造函数/赋值为noexcept / constexpr？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！