Question

我需要取十六进制数字中的所有数字并“反转”它们：全零变为非零（F），所有非零变为零。

我试过了：

void someFunction(DWORD hexVal)
{
     //...
     hexVal = ~hexVal;
     //...
}

这将0xE0000000更改为0x1FFFFFFF而不是0x0FFFFFFF。

如何产生所需的结果？

Answer 1

这应该为您提供2个字节的所需结果。你明白了4个字节。

hexval = ((hexval & 0xf000) ? 0 : 0xf000) |
         ((hexval & 0xf00) ? 0 : 0xf00) |
         ((hexval & 0xf0) ? 0 : 0xf0) |
         ((hexval & 0xf) ? 0 : 0xf);

Answer 2

假设你真的意味着你想要零 - >非零，反之亦然，逐个数字：

DWORD invertDigits(DWORD in) {
    return (
        ((in & (0xF << 28)) ? 0x0 : (0xF << 28)) |
        ((in & (0xF << 24)) ? 0x0 : (0xF << 24)) |
        ((in & (0xF << 20)) ? 0x0 : (0xF << 20)) |
        ((in & (0xF << 16)) ? 0x0 : (0xF << 16)) |
        ((in & (0xF << 12)) ? 0x0 : (0xF << 12)) |
        ((in & (0xF << 8)) ? 0x0 : (0xF << 8)) |
        ((in & (0xF << 4)) ? 0x0 : (0xF << 4)) |
        ((in & (0xF << 0)) ? 0x0 : (0xF << 0))
    );
}

Answer 3

这是按位NOT运算的期望结果。 0xE0000000 + 0x1FFFFFFF = 0xFFFFFFFF

执行所需操作的绝对最快方法是将其拆分为字节并使用查找表。

此解决方案使处理器相当于：24次添加，4次乘法和4次内存查找。乘法是数组索引的一部分。所有简单的数学运算都以大约相同的速度运行，除了稍微长一些的乘法和内存查找。您的里程可能会有所不同，具体取决于您的处理器架构和执行的编译器优化。

unsigned int transform1(unsigned int value)
{
    // static const unsigned char ZZ = 0x0, ZF = 0xF, FZ = 0xF0, FF = 0xFF; // for C++

    #define ZZ (unsigned char) 0x00
    #define FZ (unsigned char) 0xF0
    #define ZF (unsigned char) 0x0F
    #define FF (unsigned char) 0xFF

    static const unsigned char lookup[256] = 
    {
        FF, FZ, FZ, FZ, FZ, FZ, FZ, FZ, FZ, FZ, FZ, FZ, FZ, FZ, FZ, FZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
        ZF, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ, ZZ,
    }; // array takes up 1KB of RAM

    unsigned int result = 0;

    result |= lookup[(unsigned int)((value & (FF << 0 )) >> 0) ] << 0;
    result |= lookup[(unsigned int)((value & (FF << 8 )) >> 8) ] << 8;
    result |= lookup[(unsigned int)((value & (FF << 16)) >> 16)] << 16;
    result |= lookup[(unsigned int)((value & (FF << 24)) >> 24)] << 24;
    return result;
}

Answer 4

从MSB开始，您可能必须逐字节地进行操作。检查值是否在16 ^ 6和16 ^ 7之间（假设这是无符号的）。如果是，请添加到新号码0.如果不是，请添加新号码2 ^ 31 + 2 ^ 30 + 2 ^ 29 + 2 ^ 28。

看看我的目标是什么？

Answer 5

所以反转和否定是两回事。

反转取每一位并产生其补码：

 0xE0000000 = 1110 0000 0000 0000 0000 0000 0000 0000
~0xE0000000 = 0001 1111 1111 1111 1111 1111 1111 1111 = 0x1FFFFFFF

如果你想要“零以外的任何东西需要变为零”你想要布尔否定，即

hexVal = !hexVal;

编辑：好的，所以在阅读了其他一些答案之后，我终于得到了提问者的问题，这是我的个人版本，使用了一个巨大的数学表达式

n = ~(n | ((n & 0x77777777) << 1) | ((n & 0x88888888) >> 3)
        | ((n & 0x33333333) << 2) | ((n & 0xCCCCCCCC) >> 2)
        | ((n & 0x11111111) << 3) | ((n & 0xEEEEEEEE) >> 1));

反转十六进制数字的数字

5 个答案: