std::is_unsigned::价值定义良好的?-Isstd::is_unsigned::valuewelldefined?

10  

Yes, it is well-defined, as is any other unary type trait.

是的，它是明确的，就像任何其他一元类型特征一样。

C++14 (n4140) 20.10.4/2 "Unary type traits" mandates:

C+ 14 (n4140) 20.10.4/2“一元类型特征”

Each of these templates shall be a UnaryTypeTrait (20.10.1) with a BaseCharacteristic of true_type if the corresponding condition is true, otherwise false_type.

如果相应的条件为true，则每个模板都应该是一个UnaryTypeTrait(20.10.1)，如果对应的条件为true，则true_type的BaseCharacteristic为true_type，否则为false type。

20.10.1/1:

20.10.1/1:

A UnaryTypeTrait describes a property of a type. It shall be a class template that takes one template type argument and, optionally, additional arguments that help define the property being described. It shall be DefaultConstructible, CopyConstructible, and publicly and unambiguously derived, directly or indirectly, from its BaseCharacteristic, which is a specialization of the template integral_constant (20.10.3), with the arguments to the template integral_constant determined by the requirements for the particular property being described. The member names of the BaseCharacteristic shall not be hidden and shall be unambiguously available in the UnaryTypeTrait.

不arytype trait描述类型的属性。它应该是一个类模板，它接受一个模板类型参数，并且可以选择接受帮助定义所描述的属性的附加参数。它应该是默认可构造的、可复制的，并且直接或间接地、公开地、明确地从它的基属性派生出来，基属性是integral_constant模板的专门化(20.10.3)，模板integral_constant的参数由所描述的特定属性的需求决定。基性特征的成员名称不应被隐藏，并且在unarytype特征中应该是明确可用的。

From this it follows that the construct std::is_unsigned::value has to be well-defined for any type T, whether the concept of "signedness" makes sense for the type or not.

由此可以得出这样的结论:构造std::is_unsigned :::value必须为任何类型T定义良好的定义，无论“签名权”的概念对类型是否有意义。

9  

Yes it is well defined and the result should be std::is_unsigned::value == true

是的，它的定义很好，结果应该是std::is_unsigned ::value == true。

The documentation for std::is_signed says

std::is_signed说的文档

If T is a signed arithmetic type, provides the member constant value equal true. For any other type, value is false.

如果T是一个有符号的算术类型，则提供成员常量值等于true。对于任何其他类型，值都是假的。

So then if you look at std::is_arithmetic

如果你看看std: is_算术

If T is an arithmetic type (that is, an integral type or a floating-point type), provides the member constant value equal true. For any other type, value is false.

如果T是一种算术类型(即整数类型或浮点类型)，则提供成员常量值为true。对于任何其他类型，值都是假的。

Which finally leads to std::is_integral

这最终导致std::is_integral ?

Checks whether T is an integral type. Provides the member constant value which is equal to true, if T is the type bool, char, char16_t, char32_t, wchar_t, short, int, long, long long, or any implementation-defined extended integer types, including any signed, unsigned, and cv-qualified variants. Otherwise, value is equal to false.

检查T是否为整数类型。如果T是类型bool、char、char16_t、char32_t、char32_t、wchar_t、short、int、long、long、long、long、long，或者任何实现定义的扩展整数类型，包括任何已签名的、未签名的和cv限定的变体，那么提供的成员常量值就等于true。否则，值等于false。

Interestingly, there is another function std::numeric_limits::is_signed that states

有趣的是，还有一个std::numeric_limits::is_signed that states。

The value of std::numeric_limits::is_signed is true for all signed arithmetic types T and false for the unsigned types. This constant is meaningful for all specializations.

std:::numeric_limits ::is_signed为所有有符号算术类型T的真值，无符号类型为假值。这个常数对所有的专门化都有意义。

Where the specialization for bool is listed as false, which also confirms that bool is considered unsigned.

其中bool的专门化被列出为false，这也确认了bool被认为是无符号的。