位掩码 - C 中的按位运算

Question

示例：我有整数 0000010010001110

的二进制表示

如何通过 110..... 0....... 屏蔽这些位？我需要在掩码中保存零并保存所有活动位在以下整数 110010001110

我是按位运算的新手所以请给我一些想法或建议，谢谢。

更新。我需要屏蔽 wchar_t 并以 unicode (UTF-8) 格式输出代表性

Read the UTF-8 specs for more detail, but at a high level:

Code points 0 – 007F are stored as regular, single-byte ASCII. Code points 0080 and above are converted to binary and stored (encoded) in a series of bytes. The first “count” byte indicates the number of bytes for the codepoint, including the count byte. These bytes start with 11..0:

110xxxxx (The leading “11” is indicates 2 bytes in sequence, including the “count” byte)

1110xxxx (1110 -> 3 bytes in sequence)

11110xxx (11110 -> 4 bytes in sequence)

Bytes starting with 10… are “data” bytes and contain information for the codepoint. A 2-byte example looks like this

110xxxxx 10xxxxxx

Answer 1

不清楚你到底需要什么，但是如果你需要"recognize"“计数”字节和“数据”字节的类型，对于给定的例子：

1100000110100000(11000001 10100000)

到"recognize""count"字节你可以使用：

#define BIT_MASK 0X8000 //which gives---1000 0000 0000 0000

然后使用运算符&检查是否设置了位，使用counter计算设置了多少位，使用<<运算符左移（8次最大限度）。如果出现未设置的位，则中断。

  #include <stdio.h>
  #include <stdint.h>
  #define BIT_MASK        0x8000
  #define MAX_LEFT_SHIFT  8

int main(void)
{
    uint16_t exm_num = 49568;// for example 11000001 10100000 in binary
    int i,count=0;
    for(i=0;i<MAX_LEFT_SHIFT;++i){
    if (exm_num & BIT_MASK)
        ++count;
     else
         break;
     exm_num = exm_num<<1;
 }
     return 0;
}

然后您可以使用 count 的最终值来识别类型。

给定示例的输出为 2

Answer 2

i'm need to mask wchar_t and output that in unicode (UTF-8) representation

你读过UTF-8 in the Unicode standard (Section 3.9 - Unicode Encoded Forms), or RFC 3629, or even the UTF-8 documentation on Wikipedia的官方规范了吗？

他们描述了将 21 位代码点数字拆分为编码字节序列所需的算法。请注意，wchar_t 在大多数其他平台上是 16 位 (UTF-16) on Windows but is 32-bit (UTF-32)。 UTF 之间的转换相当简单，但您必须考虑 UTF 的实际含义，因为将 UTF-16 转换为 UTF-8 与将 UTF-32 转换为 UTF-8 有点不同。

简而言之，您需要这样的东西：

uint32_t codepoint = ...;
// This is the actual codepoint number, decoded from 1 or 2 wchar_t
// elements, depending on the UTF encoding of the wchar_t sequence.
// In UTF-32, the characters are the actual codepoint numbers as-is.
// In UTF-16, only the characters <= 0xFFFF are the actual codepoint
// numbers, the rest are encoded using surrogate pairs that you would
// have to decode to get the actual codepoint numbers...

uint8_t bytes[4];
int numBytes = 0;

if (codepoint <= 0x7F)
{
    bytes[0] = (uint8_t) codepoint;
    numBytes = 1;
}
else if (codepoint <= 0x7FF)
{
    bytes[0] = 0xC0 | (uint8_t) ((codepoint >> 6) & 0x1F);
    bytes[1] = 0x80 | (uint8_t) (codepoint & 0x3F);
    numBytes = 2;
}
else if (codepoint <= 0xFFFF)
{
    bytes[0] = 0xE0 | (uint8_t) ((codepoint >> 12) & 0x0F);
    bytes[1] = 0x80 | (uint8_t) ((codepoint >> 6) & 0x3F);
    bytes[2] = 0x80 | (uint8_t) (codepoint & 0x3F);
    numBytes = 3;
}
else if (codepoint <= 0x10FFFF)
{
    bytes[0] = 0xF0 | (uint8_t) ((codepoint >> 18) & 0x07);
    bytes[1] = 0x80 | (uint8_t) ((codepoint >> 12) & 0x3F);
    bytes[2] = 0x80 | (uint8_t) ((codepoint >> 6) & 0x3F);
    bytes[3] = 0x80 | (uint8_t) (codepoint & 0x3F);
    numBytes = 4;
}
else
{
    // illegal!
}

// use bytes[] up to numBytes as needed...

可以简化成这样：

uint32_t codepoint = ...; // decoded from wchar_t sequence...

uint8_t bytes[4];
int numBytes = 0;

if (codepoint <= 0x7F)
{
    bytes[0] = 0x00;
    numBytes = 1;
}
else if (codepoint <= 0x7FF)
{
    bytes[0] = 0xC0;
    numBytes = 2;
}
else if (codepoint <= 0xFFFF)
{
    bytes[0] = 0xE0;
    numBytes = 3;
}
else if (codepoint <= 0x10FFFF)
{
    bytes[0] = 0xF0;
    numBytes = 4;
}
else
{
    // illegal!
}

for(int i = 1; i < numBytes; ++i)
{
    bytes[numBytes-i] = 0x80 | (uint8_t) (codepoint & 0x3F);
    codepoint >>= 6;
}

bytes[0] |= (uint8_t) codepoint;

// use bytes[] up to numBytes as needed...

在您的示例中，0000010010001110 是十进制 1166，十六进制 0x48E。 Codepoint U+048E 以 UTF-8 编码为字节 0xD2 0x8E，例如：

0000010010001110b -> 010010b 001110b
0xC0 or 010010b -> 0xD2
0x80 or 001110b -> 0x8E

位掩码 - C 中的按位运算

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