Question

假设我有这个十六进制值：

3FFF

问题表明我必须将其分为两个字节，如下所示： 0011 1111和1111 1111

然后我必须清除每个字节中的最高有效位（MSB），根据我的理解，我必须将最左边的位更改为零，如下所示：

0011 1111和0111 1111 在HEX中是： 3F 7F

然而问题是预期结果应该是： 7F 7F

我错过了什么吗？我想知道预期的结果是错误的还是我遗漏了在清除每个字节中的MSB位时应该做的事情。

Answer 1

“预期结果”错误。

通过清除一点来获取更高的值是不可能的。^†

我没有验证你自己建议的结果，但从表面上看，这对我来说是合乎逻辑的。

_{^† inb4学生指出两个补码的符号位;去吧PLZ！}

Answer 2

使用掩码和逻辑运算：

#include <iostream>
using namespace std;

int main()
{

    unsigned char a = 255; // 11111111 // consider your variable is 255
    unsigned char LowNibble, HiNibble, tmp = a; // self-explanatory
    tmp <<= 4; // discard the high nibble
    tmp >>= 4; // storing the low nibble in tmp

    LowNibble = tmp; // assigning low nibble to LowNibble
    tmp = a;  // again for Hi nibble

    HiNibble = tmp >> 4; // discard low nibble and storing high nibble in HiNibble


    cout << "LowNibble: " << (int)LowNibble << endl; // just for checking
    cout << "HiNibble: "  << (int)HiNibble << endl;

    unsigned char mask = 119;   // 01110111 // you said discard the low nibble's highest bit and the high nibble's highest bit so we make a mask of 1s only clear the forth bit of each nibble which is produces in decimal 119 

    unsigned conc = HiNibble << 4; // assign high nibble to conc
    conc |= LowNibble; //putting low nibble in the first bits of con

    // now con contains both high and low nibbles
    conc &= mask; // clear the highest bit of both two nibbles

    cout << (int)conc << endl; //checking

   // now dividing again the byte to get what you wanted two nibbles with highest bit cleared:

    tmp = conc;
    tmp <<= 4;
    tmp >>= 4;

    LowNibble = tmp;
    tmp = conc;

    HiNibble = tmp >> 4;

    cout << "LowNibble: " << (int)LowNibble << endl;
    cout << "HiNibble: "  << (int)HiNibble << endl;


    cout << endl << endl << endl;
    return 0;
}

清除最重要的两个字节的位

2 个答案: