C++位操作

发布于 2024-11-18 04:01:07 字数 336 浏览 2 评论 0原文

我正在尝试从 UTF-8 格式中提取字符值。假设我有两个字符，我从第一个字符中提取 5 位 => 10111 和另一个字符的 6 位 => 010000

那么

ch1 = 10111;
ch2 = 010000;

我如何将它们组合成 10111010000 并将其十六进制输出为 0x5d0？我是否需要转换或者是否有更简单的方法来做到这一点，因为检查文档write 似乎能够顺序读取字符，有类似的功能吗？另外，看来我需要一个字符缓冲区，因为 10111010000 是 11 位长。有人知道该怎么做吗？

原文

I am trying to extract character value from UTF-8 format. Suppose I have two characters, and I extract 5 bits from first character => 10111 and 6 bits from another character => 010000

ch1 = 10111;
ch2 = 010000;

how would I combine them to form 10111010000 and output its hex as 0x5d0? Do I need to shift or is there an easier way to do this, because checking the documentation write appear to be able to read characters sequentially, is there a similar function like this? Also, it appears I would need a char buffer since 10111010000 is 11 bits long. Does any know how to go about this?

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忆梦 2024-11-25 04:01:07

您需要使用移位以及 | 或 |= 运算符。

unsigned int ch3 = (ch1 << 6) | ch2;
// ch3 = 0000010111010000

我在这里假设 unsigned int 是 16 位。您的里程可能会有所不同。

You need to use shifting, plus the | or |= operator.

unsigned int ch3 = (ch1 << 6) | ch2;
// ch3 = 0000010111010000

I'm assuming here that an unsigned int is 16 bits. Your mileage may vary.

回复收藏 0 原文

亣腦蒛氧 2024-11-25 04:01:07

您肯定需要使用移位和或。

首先，声明一个正确大小的无符号整数类型。我喜欢 stdint.h 中定义的 C99 类型，但您的 C++ 编译器可能没有它们。如果您没有uint16_t，那么您可以使用unsigned short。即 16 位宽，可容纳 11 位。

然后你会找出哪些位进入高位。看起来应该是：

unsigned short ch1 = 0x17;
unsigned short ch2 = 0x10;
unsigned short result = (ch1 << 6) | ch2;

You will definitely need to use shift and OR.

First, declare an unsigned integer type of the right size. I like the C99 types defined in stdint.h but your C++ compiler may not have them. If you don't have uint16_t then you can use unsigned short. That is 16 bits wide and can hold 11 bits.

Then you would figure out which bits go into the high bits. It looks like it should be:

unsigned short ch1 = 0x17;
unsigned short ch2 = 0x10;
unsigned short result = (ch1 << 6) | ch2;

回复收藏 0 原文

梦毁影碎の 2024-11-25 04:01:07

1：将它们组合在一起：

char bytes[2] = { 0x17, 0x10 }; // for example

unsigned short result = 0;      // 00000000  00000000
result = bytes[0] << 6;         // 101 11000000
result |= bytes[1];             // 101 11010000

其打印为十六进制：

std::cout << std::showbase << std::hex << <what you want to print>;

2：在本例中将

std::cout << std::showbase << std::hex << result
// output: 0x5d0 if it is little-endian, it depends on your operating system

1: for combining them together:

char bytes[2] = { 0x17, 0x10 }; // for example

unsigned short result = 0;      // 00000000  00000000
result = bytes[0] << 6;         // 101 11000000
result |= bytes[1];             // 101 11010000

2: for printing it out as hex

std::cout << std::showbase << std::hex << <what you want to print>;

in this case:

std::cout << std::showbase << std::hex << result
// output: 0x5d0 if it is little-endian, it depends on your operating system

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输什么也不输骨气 2024-11-25 04:01:07

首先，来自 K&R：“几乎所有关于位域的内容都依赖于实现”。

以下内容适用于 MS Visual Studio 2008：

#include <stdio.h>
#include <string.h>

struct bitbag {
    unsigned int ch2 : 6;
    unsigned int ch1 : 6;
};

int main ()
{
    struct bitbag bits;

    memset(&bits, 0, sizeof(bits));

    bits.ch1 = 0x17;    // 010111
    bits.ch2 = 0x10;    // 010000

    printf ("0x%06x 0x%06x\n", bits.ch1, bits.ch2);
    printf ("0x%0x\n", bits);

    return 0;
}

生成输出：

0x000017 0x000010
0x5d0

但是我无法保证它在所有编译器中都以相同的方式工作。请注意将所有填充初始化为零的 memset。

First, from K&R: "Almost everything about bitfields is implementation dependent".

The following works on MS Visual Studio 2008:

#include <stdio.h>
#include <string.h>

struct bitbag {
    unsigned int ch2 : 6;
    unsigned int ch1 : 6;
};

int main ()
{
    struct bitbag bits;

    memset(&bits, 0, sizeof(bits));

    bits.ch1 = 0x17;    // 010111
    bits.ch2 = 0x10;    // 010000

    printf ("0x%06x 0x%06x\n", bits.ch1, bits.ch2);
    printf ("0x%0x\n", bits);

    return 0;
}

Produces the output: