c++ - 连接按位运算的正确方法？

Question

我需要连接一些按位运算，但当前输出似乎是错误的。拆分操作类似于这样:

unsigned char a = 0x12
unsigned char x = 0x00;
x = a << 4;
x = x >> 4;

预期结果 x = 0x02； 当前结果 x = 0x02;

如果我尝试连接操作，结果是不正确的:

unsigned char a = 0x12
unsigned char x = 0x00;
x = (a << 4) >> 4;

预期结果 x = 0x02； 当前结果 x = 0x12;

提前感谢您的任何建议。

Answer 1

问题是(a << 4)被转换为 int (通过 Integral promotion )，所以 (0x12 << 4) >> 4本质上是 0x12

您要做的是转换回 (a << 4)至 unsigned char通过使用 static_cast

最终代码:

   unsigned char a = 0x12;
   unsigned char x = 0x00;
   x = static_cast<unsigned char>(a << 4) >> 4;