c++ - 无捕获的 lambda 是结构类型吗？

Question

P1907R1 ，为 C++20 接受，引入了结构类型，这是非类型模板参数的有效类型。
GCC 和 Clang 都接受以下 C++2a 代码片段:

template<auto v>
constexpr auto identity_v = v;

constexpr auto l1 = [](){};
constexpr auto l2 = identity_v<l1>; 

暗示无捕获 lambda 的类型是结构类型。
题

无捕获 lambda 确实满足其类型成为结构类型的要求吗？

Answer 1

除非另有明确说明，以下所有标准引用均指 N4861 (March 2020 post-Prague working draft/C++20 DIS) .

无捕获 lambda 的类型(其闭包类型)是结构类型
以后，我们将把 lambda 的类型单独称为闭包类型。
如下面的标准段落所示，无捕获 lambda 的闭包类型:

满足它是文字(类)类型的要求，而且

满足对文字类型的要求，使其成为结构类型，

并且因此可以用作非类型模板参数的类型，例如示例片段

template<auto v>
constexpr auto identity_v = v;

constexpr auto l1 = [](){};
constexpr auto l2 = identity_v<l1>;

确实是良构的。

lambda 的闭包类型是非 union 类类型
受 [expr.prim.lambda.closure]/1 管辖[ 重点 矿]

The type of a lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union class type, called the closure type, whose properties are described below.

闭包类型是非 union 类类型。

无捕获 lambda 的闭包类型是文字(类)类型
受 [basic.types]/10 管辖[提取， 重点 矿]

A type is a literal type if it is:

• [...]
• a possibly cv-qualified class type that has all of the following properties:
  • it has a constexpr destructor ([dcl.constexpr]),
  • it is either a closure type ([expr.prim.lambda.closure]), an aggregate type ([dcl.init.aggr]), or has at least one constexprconstructor or constructor template (possibly inherited from a baseclass) that is not a copy or move constructor,
  • if it is a union, at least one of its non-static data members is of non-volatile literal type, and
  • if it is not a union, all of its non-static data members and base classes are of non-volatile literal types.

闭包类型是文字类型，如果

有一个 constexpr 析构函数，如果

它的所有非静态数据成员都是非 volatile 文字类型。

无捕获 lambda 的闭包类型没有非静态数据成员，因此满足后一个要求。前者呢，一个 constexpr 析构函数？
隐式生成的 constexpr 析构函数
受 [expr.prim.lambda.closure]/14 管辖

The closure type associated with a lambda-expression has an implicitly-declared destructor ([class.dtor]).

闭包类型的析构函数是隐式声明的。此外， [/dcl.fct.def.default]/5描述 [提取物， 重点 矿]

Explicitly-defaulted functions and implicitly-declared functions are collectively called defaulted functions, and the implementation shall provide implicit definitions for them ([class.ctor], [class.dtor], [class.copy.ctor], [class.copy.assign]), [...]

集体术语默认函数还包括隐式声明的析构函数。
最后， [class.dtor]/9

A defaulted destructor is a constexpr destructor if it satisfies the requirements for a constexpr destructor ([dcl.constexpr]).

描述默认的析构函数是 constexpr 析构函数，如果它们满足 [dcl.constexpr] 的要求，特别是 [dcl.constexpr]/3和 [dcl.constexpr]/5 [提取， 重点 矿]

[dcl.constexpr]/3 The definition of a constexpr function shall satisfy the following requirements:

• [...]
• if the function is a constructor or destructor, its class shall not have any virtual base classes;
• [...]

[dcl.constexpr]/5 The definition of a constexpr destructor whose function-body is not = delete shall additionally satisfy the following requirement:

• for every subobject of class type or (possibly multi-dimensional) array thereof, that class type shall have a constexpr destructor.

所有这些都满足无捕获 lambda 的闭包类型(没有基类，也没有子对象；后者见 [intro.object]/2)。
因此，无捕获 lambda 的闭包类型是文字类型。

无捕获 lambda 的闭包类型是结构类型
根据 [temp.param]/6和 [temp.param]/7 [提取， 重点 矿]

[temp.param]/6 A non-type template-parameter shall have one of the following (possibly cv-qualified) types:

• a structural type (see below),
• [...]

[temp.param]/7

A structural type is one of the following:

• a scalar type, or
• an lvalue reference type, or
• a literal class type with the following properties:
  • all base classes and non-static data members are public and non-mutable and
  • the types of all bases classes and non-static data members are structural types or (possibly multi-dimensional) array thereof.

如果文字类类型没有基类和非静态数据成员，那么它就是结构类型。这两者都适用于无捕获 lambda，因此，无捕获 lambda 的闭包类型是结构类型。

关于允许 lambda 的闭包类型为文字类型的初衷的一些说明
N4487提议允许某些 lambda 表达式和对某些闭包对象的操作出现在常量表达式中，并包含一个专门的部分，用于讨论闭包类型是文字类型的主题:

The closure object should be a literal type if the type of each of its data­ members is a literal type.

A closure type in C++14 can never be a literal type – even if all itsdata members are literal types – because it lacks a constexprconstructor that is not a copy or move constructor. If such a closuretype was allowed to have an implicitly defined default constructor itwould be constexpr, making it a literal type. But, because closuretypes, by definition, must have their default constructors deleted,the implementation is prohibited from implicitly defining one. [...]

P0170R1包含来自 N4487 的核心措辞，已被 C++17 接受并实现。
然而此时(C++14和C++17)，析构函数不能是constexpr，因此自然不要求字面类型有constexpr析构函数； [basic.types]/10.5.1在 N4140 (C++14) 以及 [basic.types]/10.5.1 中在 N4659 (C++17) 中，相反要求析构函数是微不足道的:

A type is a literal type if it is:

• [...]
• a class type (Clause [class]) that has all of the following properties:
  • it has a trivial destructor,
  • [...]

P1907R1 ，为 C++20 接受，扩展了模板参数对象具有不断销毁的要求； [temp.param]/8 [ 重点 矿]:

An id-expression naming a non-type template-parameter of class type T denotes a static storage duration object of type const T, known as a template parameter object, whose value is that of the corresponding template argument after it has been converted to the type of the template-parameter. All such template parameters in the program of the same type with the same value denote the same template parameter object. [...] A template parameter object shall have constant destruction.

和， P0784R7 ，也被 C++20 接受，特别是引入了 constexpr 析构，包括对类型为字面类型的要求的更新；在论文的早期版本中特别描述， P0784R1 :

The proposed rules for constexpr destructors are:

• [...]
• A literal type requires a constexpr destructor (previously, the stronger requirement of a trivial destructor was made)