非类型模板参数能否为"void *"类型? [英] Can a non-type template parameter be of type "void*"?

问题描述

Y牛-亚当·内夫罗蒙特 说:

类型为void*的非类型模板参数不允许在 至少是该标准的某些版本.

Non-type template parameters of type void* are not allowed in at least some versions of the standard.

这是真的吗? 如果为真，则在哪些版本的标准中不允许类型为void*的非类型模板参数?

Is this true? If it is true, in which versions of the standard are non-type template parameters of type void* not allowed?

(注意:如 评论 回答 另一条评论， 这是关于非类型模板参数的， 不是模板类型参数， 可以是每个有效的 type-id [temp.arg.type] ， 包括void*.

(Note: as noted in a comment to answer another comment, this is about non-type template parameters, not template type arguments, which can be any valid type-id per [temp.arg.type], including void*.

推荐答案

TL; DR

类型为void*的模板参数自C ++ 20起有效. 它们在C ++ 20之前无效.

TL;DR

Template parameters of type void* are valid since C++20. They are invalid prior to C++20.

C ++ 20放宽了对非类型模板参数类型的限制， 因此，我们首先对其进行调查.

C++20 relaxed the restrictions on the type of a non-type template parameter, so let's investigate it first.

当前草案(截至UTC 2019年5月6日10:00)在

The current draft (as of UTC 10:00, May 6, 2019) says in [temp.param]/4:

非类型 template-parameter 必须具有以下之一 (可选，具有简历资格)类型:

A non-type template-parameter shall have one of the following (optionally cv-qualified) types:

• 具有强结构等式([class.compare.default])的文字类型，
• 左值引用类型
• 包含占位符类型([dcl.spec.auto])的类型，或
• 推论类类型([dcl.type.class.deduct])的占位符.

• a literal type that has strong structural equality ([class.compare.default]),
• an lvalue reference type,
• a type that contains a placeholder type ([dcl.spec.auto]), or
• a placeholder for a deduced class type ([dcl.type.class.deduct]).

void*是指针类型. 指针类型是标量类型( [基本.types]/9 ). 标量类型是文字类型(

void* is a pointer type. A pointer type is a scalar type ([basic.types]/9). A scalar type is a literal type ([basic.types]/10). Therefore, void* is a literal type. The first bullet is the relevant one.

进一步跟踪， [class.compare.default]/3 说:

如果给定的全值x为，则类型C具有强烈的结构相等性 键入const C，或者:

A type C has strong structural equality if, given a glvalue x of type const C, either:

• C是非类类型，x <=> x是类型std::strong_ordering或std::strong_equality或

• C is a non-class type and x <=> x is a valid expression of type std::strong_ordering or std::strong_equality, or

C是具有==运算符的类类型，该运算符在C的定义中默认定义，x == x在上下文中格式正确 转换为bool的所有C基类子对象，并且都是非静态的 数据成员具有很强的结构相等性，并且C没有mutable 或volatile子对象.

C is a class type with an == operator defined as defaulted in the definition of C, x == x is well-formed when contextually converted to bool, all of C's base class subobjects and non-static data members have strong structural equality, and C has no mutable or volatile subobjects.

void*是非类类型， 所以第一个项目符号是相关的. 现在，问题归结为x <=> x的类型 其中，x是类型为void* const的glvalue(不是const void*). 根据 [expr.spaceship]/8 :

void* is a non-class type, so the first bullet is relevant. Now the question boils down to the type of x <=> x where x is a glvalue of type void* const (not const void*). Per [expr.spaceship]/8:

如果复合指针类型为对象指针类型，则p <=> q为 类型为std::strong_­ordering.如果两个指针操作数p和q 比较等于([expr.eq])，p <=> q收益 std::strong_­ordering::equal;如果p和q比较不相等，则p <=> q得出std::strong_­ordering::less如果q比较大于 p和std::strong_­ordering::greater，如果p比较大于 q([expr.rel]).否则，结果将不确定.

If the composite pointer type is an object pointer type, p <=> q is of type std::strong_­ordering. If two pointer operands p and q compare equal ([expr.eq]), p <=> q yields std::strong_­ordering::equal; if p and q compare unequal, p <=> q yields std::strong_­ordering::less if q compares greater than p and std::strong_­ordering::greater if p compares greater than q ([expr.rel]). Otherwise, the result is unspecified.

请注意，void*是对象指针类型((

Note that void* is an object pointer type ([basic.compound]/3). Therefore, x <=> x is of type std::strong_ordering. Thus the type void* has strong structural equality.

因此，在当前的C ++ 20草案中， 允许将void*作为模板参数类型的类型.

Therefore, in the current C++20 draft, void* is allowed as the type of a template parameter type.

现在，我们解决C ++ 17问题. [temp.param] 说:

Now we address C++17. [temp.param] says:

非类型 template-parameter 必须具有以下之一 (可选，具有简历资格)类型:

A non-type template-parameter shall have one of the following (optionally cv-qualified) types:

• 整数或枚举类型
• 指向对象的指针或指向函数的指针，
• 对对象的左值引用或对函数的左值引用，
• 成员的指针，
• std​::​nullptr_­t或
• 包含占位符类型的类型.

• integral or enumeration type,
• pointer to object or pointer to function,
• lvalue reference to object or lvalue reference to function,
• pointer to member,
• std​::​nullptr_­t, or
• a type that contains a placeholder type.

请注意，指向对象的指针"不包含void* 每个 [basic.compound]/3 :

Note that "pointer to object" doesn't include void* per [basic.compound]/3:

[注意:"指向void的指针没有指针到对象的类型，因为void不是对象类型. — 尾注]

[ Note: A pointer to void does not have a pointer-to-object type, however, because void is not an object type. — end note ]

以上六个项目符号均不包含void* 作为模板参数的可能类型. 因此，在C ++ 17中， 模板参数的类型不得为void*.

None of the above six bullets include void* as a possible type of a template parameter. Therefore, in C++17, a template parameter shall not have type void*.

C ++ 11和C ++ 14的措辞相同 除了不存在有关占位符类型的项目符号. 一般来说， 在C ++ 20之前， 模板参数的类型不得为void*.

The wording is the same for C++11 and C++14 except that the bullet about placeholder types are not there. In general, prior to C++20, a template parameter shall not have type void*.

T.C. 在 评论 没有人诊断出这个IHRC. 让我们测试一下编译器是否在C ++ 17模式下进行诊断 并显示以下最小示例:

T.C. says in a comment that nobody diagnoses this IHRC. Let's test whether compilers diagnose that in C++17 mode with the minimal example shown below:

template <void*>
class C {};

int main()
{
    C<nullptr> x;
    (void) x;
}

代码可以在以下位置编译并正常运行 GCC 9.1.0 ， GCC 8.3.0 ， GCC 7.3.0 ， GCC 6.3.0 ， GCC 5.5.0 ， C 8.0.0 ， C 7.0.0 ， C 6.0.1 ， 和 C 5.0.0 .

The code compiles and runs fine on GCC 9.1.0, GCC 8.3.0, GCC 7.3.0, GCC 6.3.0, GCC 5.5.0, Clang 8.0.0, Clang 7.0.0, Clang 6.0.1, and Clang 5.0.0.

NathanOliver 在

NathanOliver told me in a comment that someone told him some compilers will error, but the major ones don't. Therefore, as far as I am able to confirm here, T.C.'s statement is correct — nobody diagnoses this.

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