我正在研究一个问题，我必须使用位操作来计算给定数字的五分之八（5/8）吗？
对于正数，我可以很容易地做到。基本上，它是( (x << 2) + x )>> 3。
然而，对于负数，它似乎不起作用。我环顾网络，显然，我必须增加一个因子7，但是，我不能安静地看看为什么需要这个？
答案 0 :(得分：2)
使用轮向负无穷大进行分割，而正常C分组向零舍入。
也就是说，-9 / 8 == 1（即-1.25向零舍入为-1），但是-9 >> 2 == -2（即，`-1.25'向负无穷大舍入为 - 2）。
要解决这个问题，对于除以8 的特定情况，您可以在负数的情况下添加7以“调整”被除数，以便按预期进行舍入。
这个问题的全部内容假设您的C编译器为签名的右移实现了“算术右移”。几乎所有的架构/编译器组合都有，但标准并不保证。
答案 1 :(得分：1)
对于正面x，x >> 3和x / 8都朝向零。
对于否定x，x >> 3向负无穷大舍入，而x / 8向零舍入。例子：
-1 >> 3 = -1 -1 / 8 = 0 different
-2 >> 3 = -1 -2 / 8 = 0 different
-3 >> 3 = -1 -3 / 8 = 0 different
-4 >> 3 = -1 -4 / 8 = 0 different
-5 >> 3 = -1 -5 / 8 = 0 different
-6 >> 3 = -1 -6 / 8 = 0 different
-7 >> 3 = -1 -7 / 8 = 0 different
-8 >> 3 = -1 -8 / 8 = -1 same
-9 >> 3 = -2 -9 / 8 = -1 different
-10 >> 3 = -2 -10 / 8 = -1 different
-11 >> 3 = -2 -11 / 8 = -1 different
-12 >> 3 = -2 -12 / 8 = -1 different
-13 >> 3 = -2 -13 / 8 = -1 different
-14 >> 3 = -2 -14 / 8 = -1 different
-15 >> 3 = -2 -15 / 8 = -1 different
-16 >> 3 = -2 -16 / 8 = -2 same
-17 >> 3 = -3 -17 / 8 = -2 different
-18 >> 3 = -3 -18 / 8 = -2 different
-19 >> 3 = -3 -19 / 8 = -2 different
当分子（x）是分母（8）的倍数时，结果是相同的。对于其他7/8的结果，结果相差1.这意味着如果我们希望>> 3的行为与/ 8相同，我们需要更改分子。
一般来说，如果你有一个向下舍入的整数除法运算符，你可以通过在分子中加上（分母-1）来使它向上舍入。但是，让我们按照婴儿步骤到达那里。假设我们通过添加1来改变分子：
( -1 + 1) >> 3 = 0 -1 / 8 = 0 same
( -2 + 1) >> 3 = -1 -2 / 8 = 0 different
( -3 + 1) >> 3 = -1 -3 / 8 = 0 different
( -4 + 1) >> 3 = -1 -4 / 8 = 0 different
( -5 + 1) >> 3 = -1 -5 / 8 = 0 different
( -6 + 1) >> 3 = -1 -6 / 8 = 0 different
( -7 + 1) >> 3 = -1 -7 / 8 = 0 different
( -8 + 1) >> 3 = -1 -8 / 8 = -1 same
( -9 + 1) >> 3 = -1 -9 / 8 = -1 same
(-10 + 1) >> 3 = -2 -10 / 8 = -1 different
(-11 + 1) >> 3 = -2 -11 / 8 = -1 different
(-12 + 1) >> 3 = -2 -12 / 8 = -1 different
(-13 + 1) >> 3 = -2 -13 / 8 = -1 different
(-14 + 1) >> 3 = -2 -14 / 8 = -1 different
(-15 + 1) >> 3 = -2 -15 / 8 = -1 different
(-16 + 1) >> 3 = -2 -16 / 8 = -2 same
(-17 + 1) >> 3 = -2 -17 / 8 = -2 same
(-18 + 1) >> 3 = -3 -18 / 8 = -2 different
(-19 + 1) >> 3 = -3 -19 / 8 = -2 different
现在我们有2/8的结果匹配。尝试添加2：
( -1 + 2) >> 3 = 0 -1 / 8 = 0 same
( -2 + 2) >> 3 = 0 -2 / 8 = 0 same
( -3 + 2) >> 3 = -1 -3 / 8 = 0 different
( -4 + 2) >> 3 = -1 -4 / 8 = 0 different
( -5 + 2) >> 3 = -1 -5 / 8 = 0 different
( -6 + 2) >> 3 = -1 -6 / 8 = 0 different
( -7 + 2) >> 3 = -1 -7 / 8 = 0 different
( -8 + 2) >> 3 = -1 -8 / 8 = -1 same
( -9 + 2) >> 3 = -1 -9 / 8 = -1 same
(-10 + 2) >> 3 = -1 -10 / 8 = -1 same
(-11 + 2) >> 3 = -2 -11 / 8 = -1 different
(-12 + 2) >> 3 = -2 -12 / 8 = -1 different
(-13 + 2) >> 3 = -2 -13 / 8 = -1 different
(-14 + 2) >> 3 = -2 -14 / 8 = -1 different
(-15 + 2) >> 3 = -2 -15 / 8 = -1 different
(-16 + 2) >> 3 = -2 -16 / 8 = -2 same
(-17 + 2) >> 3 = -2 -17 / 8 = -2 same
(-18 + 2) >> 3 = -2 -18 / 8 = -2 same
(-19 + 2) >> 3 = -3 -19 / 8 = -2 different
显然，如果我们计算(x + i) >> 3，结果的（i + 1）/ 8匹配。因此，为了使所有结果匹配，我们为i求解（i + 1）/ 8 = 1，得到i = 7.如果我们将7加到分子上，这就是我们得到的结果：
( -1 + 7) >> 3 = 0 -1 / 8 = 0 same
( -2 + 7) >> 3 = 0 -2 / 8 = 0 same
( -3 + 7) >> 3 = 0 -3 / 8 = 0 same
( -4 + 7) >> 3 = 0 -4 / 8 = 0 same
( -5 + 7) >> 3 = 0 -5 / 8 = 0 same
( -6 + 7) >> 3 = 0 -6 / 8 = 0 same
( -7 + 7) >> 3 = 0 -7 / 8 = 0 same
( -8 + 7) >> 3 = -1 -8 / 8 = -1 same
( -9 + 7) >> 3 = -1 -9 / 8 = -1 same
(-10 + 7) >> 3 = -1 -10 / 8 = -1 same
(-11 + 7) >> 3 = -1 -11 / 8 = -1 same
(-12 + 7) >> 3 = -1 -12 / 8 = -1 same
(-13 + 7) >> 3 = -1 -13 / 8 = -1 same
(-14 + 7) >> 3 = -1 -14 / 8 = -1 same
(-15 + 7) >> 3 = -1 -15 / 8 = -1 same
(-16 + 7) >> 3 = -2 -16 / 8 = -2 same
(-17 + 7) >> 3 = -2 -17 / 8 = -2 same
(-18 + 7) >> 3 = -2 -18 / 8 = -2 same
(-19 + 7) >> 3 = -2 -19 / 8 = -2 same
答案 2 :(得分：0)
为了让它更具视觉效果，
取数字，比方说，-42（这不是5的倍数，但无论如何）
11010110
将一个小数点放在右边3个位置（这除以8）
11010.110 = -5.25
如何向上舍入：向所有分数位添加1（意味着iff不是全部为零，加法将进入整数部分），所以0.111，然后斩：
11010.110
0.111 = 7/8
--------- +
11011.101 = -4.375
chop:
11011.000 = -5
要转换为普通整数，请移至小数点恰好在最低有效位之前：
11011.000 >>s 3 =
11111011. = still -5, but in normal integer format
在代码中你只是将小数点放在虚拟中（所以什么都不做，但是就好像它在那里一样）并且分数是隐含的（所以7/8写成7）。因为正确的移位无论如何都会抛出这些位，所以斩断是不必要的。剩下的就是加7，然后换班。