需要帮助来了解"getbits()"； K& R C第2章中的方法 [英] Need help understanding "getbits()" method in Chapter 2 of K&R C

问题描述

在第2章按位运算符"一节(第2.9节)中，我很难理解其中一种示例方法的工作原理.

In chapter 2, the section on bitwise operators (section 2.9), I'm having trouble understanding how one of the sample methods works.

以下是提供的方法:

unsigned int getbits(unsigned int x, int p, int n) {
    return (x >> (p + 1 - n)) & ~(~0 << n);
}

这个想法是，对于给定的数字 x ，它将从位置 p 开始从右边开始返回 n 位(最右边的位是位置0).给出以下main()方法:

The idea is that, for the given number x, it will return the n bits starting at position p, counting from the right (with the farthest right bit being position 0). Given the following main() method:

int main(void) {
    int x = 0xF994, p = 4, n = 3;
    int z = getbits(x, p, n);
    printf("getbits(%u (%x), %d, %d) = %u (%X)\n", x, x, p, n, z, z);

    return 0;
}

输出为:

getbits(63892 (f994), 4, 3) = 5 (5)

我得到了一部分，但在大局"方面遇到了麻烦，主要是因为我不了解的小部分(没有双关语).

I get portions of this, but am having trouble with the "big picture," mostly because of the bits (no pun intended) that I don't understand.

我专门遇到问题的部分是补编:~(~0 << n).我想我得到了第一部分，涉及到 x ；我正在努力的就是这部分(然后是遮罩)，以及如何将它们全部组合在一起以真正检索这些位. (通过代码和使用calc.exe检查我的结果，我已经验证了它的作用-谢谢上帝，它具有二进制视图！)

The part I'm specifically having issues with is the complements piece: ~(~0 << n). I think I get the first part, dealing with x; it's this part (and then the mask) that I'm struggling with -- and how it all comes together to actually retrieve those bits. (Which I've verified it is doing, both with code and checking my results using calc.exe -- thank God it has a binary view!)

有帮助吗?

推荐答案

在我们的示例中，我们使用16位.在这种情况下，〜0等于

Let's use 16 bits for our example. In that case, ~0 is equal to

1111111111111111

当我们将此n位左移(在您的情况下为3)时，我们得到:

When we left-shift this n bits (3 in your case), we get:

1111111111111000

因为左侧的1被丢弃，而0则在右侧被送入.然后对其进行补充，即可得到:

because the 1s at the left are discarded and 0s are fed in at the right. Then re-complementing it gives:

0000000000000111

因此，这是在数字的最低有效部分中获取n 1位的聪明方法.

so it's just a clever way to get n 1-bits in the least significant part of the number.

您描述的"x位"已将给定数字(f994)右移了足够远，因此，最低3位是您想要的.在此示例中，您所请求的位用."括起来.字符.

The "x bit" you describe has shifted the given number (f994) right far enough so that the least significant 3 bits are the ones you want. In this example, the bits you're requesting are surrounded by '.' characters.

ff94             11111111100.101.00  # original number
>> p+1-n     [2] 0011111111100.101.  # shift desired bits to right
& ~(~0 << n) [7] 0000000000000.101.  # clear all the other (left) bits

在那里有你的东西. da ！！

And there you have your bits. Ta da !!

这篇关于需要帮助来了解"getbits()"； K& R C第2章中的方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！