c++ - 无符号位域的溢出是否保证回绕？

Question

详情

reference for bit fields at cppreference给出以下示例:

#include <iostream>
struct S {
 // three-bit unsigned field,
 // allowed values are 0...7
 unsigned int b : 3;
};
int main()
{
    S s = {7};
    ++s.b; // unsigned overflow (guaranteed wrap-around)
    std::cout << s.b << '\n'; // output: 0
}

强调有保证的环绕评论。

然而，WG21 CWG Issue 1816描述一些可能存在的位字段值规范不明确的问题，以及 [expr.post.incr]/1在最新的标准草案状态:

The value of a postfix ++ expression is the value of its operand. ...

If the operand is a bit-field that cannot represent the incremented value, the resulting value of the bit-field is implementation-defined.

但是，我不确定这是否也适用于无符号位字段的回绕。

问题

• 是否保证无符号位域溢出？

Answer 1

两者都是[expr.pos]/1和 [expr.ass]/6同意(有符号或无符号)位域上的整数溢出是实现定义的。

[expr.pos]/1

[...] If the operand is a bit-field that cannot represent the incremented value, the resulting value of the bit-field is implementation-defined.

[expr.ass]/6

When the left operand of an assignment operator is a bit-field that cannot represent the value of the expression, the resulting value of the bit-field is implementation-defined.

我已经修复了 cppreference 页面。感谢您的关注。