C 14中的有符号位字段

我相信,直到C 14,声明为int的结构的位字段仍然被解释为有符号或无符号,解释是实现定义的.参考： http://en.cppreference.com/w/cpp/language/bit_field.

在C 14中仍然如此吗？即,下面的代码是否保证按照有意义工作？

#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
}

解决方法

根据C 11标准§9.6/ p3比特字段[class.bit](强调矿)：

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.实际上,声明为signed的结构的位字段仍然被解释为有符号或无符号,解释是实现定义的.

正如@T.C年的评论中已经提到的那样.提到该问题的缺陷报告是在DR739,DR675进行的.导致C 14标准中的以下决议：

措辞“它是实现定义的是一个普通(既没有显式签名也没有未签名)char,short,int,long或long long位字段是有符号还是无符号.”,已被删除,C 14措辞现在是：

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
]

同样在§C.1.8第9章：类[diff.class]中添加了以下部分：

9.6

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

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.

Effect on original feature: The choice is implementation-defined in C,but not so in C++.

Difficulty of converting: Syntactic transformation.

How widely used: Seldom.

因此,在C 14中,对plain int类型的位字段进行了签名,并且保证发布的代码按预期工作.

原文地址：https://www.jb51.cc/c/119246.html