K＆安培; R第：需要帮助理解和QUOT; getbits（）＆QUOT;方法2章 [英] K & R Question: Need help understanding "getbits()" method in Chapter 2
问题描述
正如我前面提到的，我经历K＆安培; R，和整体在做没事吧。然而，在第2章（第2.9节）对位运算符节，我无法理解怎么样的方法之一工作 - 作为一个结果，我在与相关的练习麻烦。
As I've mentioned before, I'm going through K&R, and overall am doing all right with it. However, in chapter 2, the section on bitwise operators (section 2.9), I'm having trouble understanding how one of the sample methods works -- and as a result, I'm having trouble with the associated exercises.
（这不是对位移我以前问题的欺骗，顺便说一句，这是新和改善！）
(This isn't a dupe of my prior question on bit-shifting, btw. This is new-and-improved!)
所以这里提供的方法：
unsigned int getbits(unsigned int x, int p, int n) {
return (x >> (p + 1 - n)) & ~(~0 << n);
}
我们的想法是，对于给定数量的 X 的，它会返回的 N 的起始位置比特的 P 的，从右边计数（与最右边位位置为0）。考虑下面的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;
}
输出是：
getbits（63892（f994），4，3）= 5（5）
getbits(63892 (f994), 4, 3) = 5 (5)
我得到这个部分，但我有难同大画面，因为位的大多是（没有双关语意），我不明白。
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.
（请注意，我的不的寻找与相关的练习帮助 - 我想如果我得到这个，我能够得到这些信息）
(Note that I'm not looking for help with the associated exercises -- I think if I "got" this, I'd be able to get those.)
我专门遇到问题的部分是补充片：〜（〜0℃;＆LT; N）。我觉得我得到的第一部分，处理的 X 的;它的这一部分（然后蒙版）我正在挣扎 - 如何这一切都在一起，其实检索这些位。 （我已经验证它是干什么的，都与code和检查使用的calc.exe我的结果 - ！感谢上帝，它有一个二进制视图）
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!)
任何帮助吗？
推荐答案
让我们用16位为我们的榜样。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
当我们左移（在你的案件3）本 N 位，我们得到：
When we left-shift this n bits (3 in your case), we get:
1111111111111000
因为 1 ■在左侧被丢弃和 0 ，则在适当的加入。然后重新补充它给：
because the 1s at the left are discarded and 0s are fed in at the right. Then re-complementing it gives:
0000000000000111
所以它只是一个聪明的方式来获得 N 的数量最少显著的部分1位。
so it's just a clever way to get n 1-bits in the least significant part of the number.
在X位你描述已经转移给定的数字（f994）右远足以使最不显著3位是你想要的。在这个例子中，你请求位被包围'。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
和你有你的位。ただ！
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