在C ++ 14中的有符号位字段 [英] Signed bit field in C++14

问题描述

我相信，直到C ++ 14，声明为 int 的结构的位字段仍然被解释为 signed 或 unsigned ，解释是实现定义的。参考： http://en.cppreference.com/w/cpp/language/bit_field。

I believe that until C++14 a bit field of a struct declared as int was still interpreted as either signed or unsigned, the interpretation being implementation defined. Reference: http://en.cppreference.com/w/cpp/language/bit_field.

在C ++ 14中还是这样吗？也就是说，下面的代码是否保证可以正常工作？

Is this still the case in C++14? I.e., is the code below guaranteed to work as inteded?

#include <iostream>

struct X
{
    int f:3;
};

int main() 
{
    X x;
    x.f = -2; // is this going to be indeed signed? It seems so.
    std::cout << x.f << std::endl; // displays -2
}

推荐答案

到C ++ 11标准§9.6/ p3位字段[class.bit] （强调Mine ）：

According to C++11 standard §9.6/p3 Bit-fields [class.bit] (Emphasis Mine):

位字段不应该是静态成员。位字段必须具有
整数或枚举类型（3.9.1）。 它是实现定义的
是否是一个简单（既不明确签名也不是无符号） char ， short ，
int ， long 或 long long 位字段是有符号或无符号。 A bool value
可以成功存储在任何非零大小的位字段中。
地址的运算符&不应用于位字段，因此
不是位字段的指针。非const引用不应将
绑定到位字段（8.5.3）。 [注意：如果
引用的初始化器类型为const T&是引用位字段的左值，引用
绑定到临时初始化以保存
位字段的值;该引用不直接绑定到位域。请参阅
8.5.3。 - end note]

A bit-field shall not be a static member. A bit-field shall have integral or enumeration type (3.9.1). It is implementation-defined whether a plain (neither explicitly signed nor unsigned) char, short, int, long, or long long bit-field is signed or unsigned. A bool value can successfully be stored in a bit-field of any nonzero size. The address-of operator & shall not be applied to a bit-field, so there are no pointers to bitfields. A non-const reference shall not be bound to a bit-field (8.5.3). [ Note: If the initializer for a reference of type const T& is an lvalue that refers to a bit-field, the reference is bound to a temporary initialized to hold the value of the bit-field; the reference is not bound to the bit-field directly. See 8.5.3. —end note ]

所以你对第一部分是正确的。事实上，直到C ++ 14，声明为的结构的位域仍然被解释为 signed 或<$

So you're correct for the first part. Indeed until C++14 a bit field of a struct declared as signed was still interpreted as either signed or unsigned, the interpretation being implementation defined.

如本评论中提到的

As already mentioned in this comments by @T.C. Defect reports referring to the issue were made DR739, DR675. Resulting in the following resolutions in C++14 standard:

措辞是由实现定义的，无论是明文签名还是无签名 char ， short ， int ， long 或 long long 位字段是有符号或无符号。，被删除，现在的C ++ 14措辞是：

The wording "It is implementation-defined whether a plain (neither explicitly signed nor unsigned) char, short, int, long, or long long bit-field is signed or unsigned.", was removed, and the C++14 wording now is:

位字段不应该是静态成员。位字段必须具有
整数或枚举类型（3.9.1）。 bool值可以成功地存储在任何非零大小的位字段中的
。运营商和运营商的地址。
不应用于位字段，因此没有指向
位字段的指针。非const引用不应该绑定到位字段
（8.5.3）。 [注意：如果const T&
是一个引用位字段的左值，引用绑定到一个
临时初始化来保存位字段的值;
引用不直接绑定到位域。见8.5.3。 -end note
]

A bit-field shall not be a static member. A bit-field shall have integral or enumeration type (3.9.1). A bool value can successfully be stored in a bit-field of any nonzero size. The address-of operator & shall not be applied to a bit-field, so there are no pointers to bit-fields. A non-const reference shall not be bound to a bit-field (8.5.3). [ Note: If the initializer for a reference of type const T& is an lvalue that refers to a bit-field, the reference is bound to a temporary initialized to hold the value of the bit-field; the reference is not bound to the bit-field directly. See 8.5.3. —end note ]

也在§C.1.8第9条：classes [diff.class] 添加了以下部分：

9.6

9.6

更改：类型为plain int的位字段已签名。

Change: Bit-fields of type plain int are signed.

基本原理
模板特殊化的不一致定义。为了一致性，对于非依赖的
类型，也消除了实现自由。

Rationale: Leaving the choice of signedness to implementations could lead to inconsistent definitions of template specializations. For consistency, the implementation freedom was eliminated for non-dependent types, too.

对原始特征的影响：

转换难度：语法转换。

使用范围：很少。

strong>类型为 int 的位字段已签名，并且发布的代码可保证按预期工作。

Consequently, in C++14 bit-fields of type plain int are signed and the code posted is guaranteed to work as intended.

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