如何破解这个位域？

发布于 2024-10-19 16:09:34 字数 685 浏览 1 评论 0原文

我必须编写一个函数 setbits(x,p,n,y) ，它返回 x 以及从位置 n 位开始>p 设置为无符号字符变量 y 最右边的 n 位（保持其他位不变）。

例如，如果x = 10101010（十进制170）和y = 10100111（十进制167）和n = 3和p = 6< /code> 假设你需要去掉 3 位 y (111) 并将它们放入 x 的位置 10xxx010 才能得到答案 <代码>10111010。

该函数应以二进制形式打印结果。

输出应该是这样的：

x = 10101010 (binary)
y = 10100111 (binary)
setbits n = 3, p = 6 gives x = 10111010 (binary)

嗨，我在位字段中遇到了这个问题。
我该怎么办？

Typedef struct {
    unsigned char x:8;
    unsigned char y:8;
} var;

原文

I have to write a function setbits(x,p,n,y) who returns x with the n bits that begin at position p set to the right most n bits of an unsigned char variable y (leaving other bits unchanged).

E.g. if x = 10101010 (170 decimal) and y = 10100111 (167 decimal) and n = 3 and p = 6 say then you need to strip off 3 bits of y (111) and put them in x at position 10xxx010 to get answer 10111010.

This function should print out the result in binary form.

The output should be like this :

x = 10101010 (binary)
y = 10100111 (binary)
setbits n = 3, p = 6 gives x = 10111010 (binary)

Hi I came across this problem in bit field.

How should I go about it?

Typedef struct {
    unsigned char x:8;
    unsigned char y:8;
} var;

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随梦而飞# 2024-10-26 16:09:34

位域是一个特定的 C 发明，与此无关。例如，您可以使用

struct RGB16 {
   unsigned int R : 5;
   unsigned int G : 6;
   unsigned int B : 5;
};

16 位结构，其中包含 3 个组件，每个组件都小于一个字节。然而，C 标准没有提及它们的相对位置。您不能假设它们在内存中的布局为 RRRRRGGGGGGBBBBB。此外，位字段宽度是恒定的，在您的情况下，所需的位是可变的。因此，谷歌搜索“C 位字段”不会帮助您解决这个问题。

至于你的问题，你应该知道 x & binary(11000011) 将中间 4 位清零，并且 y | binary(00111100) 将中间 4 位设置为 1。

Bitfields are a specific C invention that have nothing to do with this. For instance, you could have

struct RGB16 {
   unsigned int R : 5;
   unsigned int G : 6;
   unsigned int B : 5;
};

which is a 16 bits structure, with 3 components that are each smaller than a byte. However, the C standard say nothing about their relative position. You can't assume that they're laid out in memory as RRRRRGGGGGGBBBBB. Furthermore, the bitfieldwidths are constant, and in your case the bits needed are variable. So, googling for "C bitfields" won't help you with this problem.

As for your problem, you should know that x & binary(11000011) zeroes out the middle 4 bits, and y | binary(00111100) sets the middle 4 bits to one.

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