c++ - C++中位运算符的定义是什么？

Question

根据标准，运算符 << 为第一个负符号操作数产生未定义行为。

C++11 5.8.2

The value of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are zero-
filled. If E1 has an unsigned type, the value of the result is E1 × 2 pow E2,
reduced modulo one more than the maximum value representable in the result type.
Otherwise, if E1 has a signed type and non-negative value, and E1 × 2 pow E2 is
representable in the result type, then that is the resulting value; otherwise,
the behavior is undefined

这是可以理解的，因为整数在内存中的布局是实现定义的。

C++11 3.9.1.7

this International Standard permits 2’s complement, 1’s complement and
signed magnitude representations for integral types.

另一方面，标准似乎没有明确定义按位 & |和 ^ 应该做。

C++11 5.11 按位与运算符

and-expression:
    equality-expression
    and-expression & equality-expression
1 The usual arithmetic conversions are performed; the result is the bitwise 
AND function of the operands. The operator applies only to integral
or unscoped enumeration operands.

C++11 5.12 按位异或运算符

exclusive-or-expression:
    and-expression
    exclusive-or-expression ˆ and-expression
1 The usual arithmetic conversions are performed; the result is the bitwise
exclusive OR function of the operands. The operator applies only to integral
or unscoped enumeration operands.

C++11 5.13 按位包含或运算符

inclusive-or-expression:
    exclusive-or-expression
    inclusive-or-expression | exclusive-or-expression
1 The usual arithmetic conversions are performed; the result is the bitwise
inclusive OR function of its operands. The operator applies only to integral
or unscoped enumeration operands.

这些运算符的定义完全让我难以理解。它在标准的其他地方吗？是否为有符号整数定义了结果实现？

作为示例，让我们看一下这段代码:

signed char a=-1;
signed char b=3;
signed char c=a&b;

加上2的补码，a为1111 1111，b为0000 0011，最后c等于0000 0011(+3)。

用1的补码，a是1111 1110，b是0000 0011，c等于0000 0010(+2)吗？

带符号数，a为1000 0001，b为0000 0011，c是否等于0000 0001(+1)？

如果您可以访问使用 1 的补码或符号幅度的平台，那么在这些平台上的结果是什么？

Answer 1

按位运算独立地对每个位进行操作，无论每个位在被解释为数字类型的一部分时可能意味着什么。

所以是的，10000001 & 00000011 == 00000001，无论每个位代表一个符号还是值的一部分。