我知道对于无符号整数，如果除数是2的幂，我可以用位掩码替换模运算。浮动数字是否有类似的属性？也就是说，是否有任何数字n可以比一般情况更有效地计算f mod n，而不一定使用位掩码？
当然，除了一个。 脑功能
编辑：澄清一下，f是任何浮点数（在运行时确定），
n是任何格式的编译时常数，我希望结果是浮点数。
答案 0 :(得分：3)
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或n == 1.0，您可以执行以下操作：
n == -1.0在x86_64上，trunc通常会使用r = f - trunc(f);
指令，因此速度非常快。
如果ROUNDSD是2的幂，其幅度大于或等于1，并且您的平台具有本地的fma函数（对于Intel，这意味着Haswell或更新），那么您可以吗
n任何合理的编译器都应该将除法切换为乘法，并将否定值折叠为适当的FMA（或常量），从而产生乘法，截断和FMA。
这也适用于较小的2的幂，只要结果不会溢出（因此编译器不能自由替换它）。
任何编译器是否真的会这样做是另一回事。浮点余数函数使用不多，并且没有得到编译器编写者的太多关注，例如： https://bugs.llvm.org/show_bug.cgi?id=3359
答案 1 :(得分：2)
浮点类型的数学运算与整数类型的运算方式相同：如果 n 是基数的幂（二进制为2），那么 f modulo n 可以通过将表示值 n 或更大的数字（也称为高位或高位）归零来计算。
因此，对于b 15 b 14 b 13 b 12 b 的二进制整数> 11 b 10 b 9 b 8 b 7 b 6 b 5 b 4 b 3 b 2 b 1 b 0 ，我们可以简单地通过将b 15 设置为b 2 为零来计算模4的残差，只留下b 1 b <子> 0
类似地，如果浮点格式的基数是2，我们可以通过删除值为4或更大的所有数字来计算模4的残差。这不需要除法，但它确实需要检查表示值的位。仅仅一个简单的位掩码是不够的。
C标准将浮点类型表征为符号（±1），基数 b ，指数和一些基数 b 数字。因此，如果我们知道特定C实现用于表示浮点类型的格式（符号，指数和数字被编码为位的方式），则计算 f 模数的算法< em> n ，其中 n 是 b 的力量，是：
一些注意事项：
示例代码：
// This code assumes double is IEEE 754 basic 64-bit binary floating-point.
#include <math.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
// Return the bits representable double x.
static uint64_t Bits(double x)
{ return (union { double d; uint64_t u; }) { x } .u; }
// Return the double represented by bits x.
static double Double(uint64_t x)
{ return (union { uint64_t u; double d; }) { x } .d; }
// Return x modulo 2**E.
static double Mod(double x, int E)
{
uint64_t b = Bits(x);
int e = b >> 52 & 0x7ff;
// If x is a NaN, return it.
if (x != x) return x;
// Is x is infinite, return a NaN.
if (!isfinite(x)) return NAN;
// If x is subnormal, adjust its exponent.
if (e == 0) e = 1;
// Remove the encoding bias from e and get the difference in exponents.
e = (e-1023) - E;
// Calculate number of bits to keep. (Could be consolidated above, kept for illustration.)
e = 52 - e;
if (e <= 0) return 0;
if (53 <= e) return x;
// Remove the low e bits (temporarily).
b = b >> e << e;
/* Convert b to a double and subtract the bits we left in it from the
original number, thus leaving the bits that were removed from b.
*/
return x - Double(b);
}
static void Try(double x, int E)
{
double expected = fmod(x, scalb(1, E));
double observed = Mod(x, E);
if (expected == observed)
printf("Mod(%a, %d) = %a.\n", x, E, observed);
else
{
printf("Error, Mod(%g, %d) = %g, but expected %g.\n",
x, E, observed, expected);
exit(EXIT_FAILURE);
}
}
int main(void)
{
double n = 4;
// Calculate the base-two logarithm of n.
int E;
frexp(n, &E);
E -= 1;
Try(7, E);
Try(0x1p53 + 2, E);
Try(0x1p53 + 6, E);
Try(3.75, E);
Try(-7, E);
Try(0x1p-1049, E);
}
答案 2 :(得分：0)
一般来说，没有。然而，考虑到浮点的“模糊”性质（至少是IEEE 754），存在 abuses 技巧，可以通过近似从根本上加速某些计算，但代价是内存，精度，可移植性，可维护性，或这些的组合。
最简单的方法是预先计算操作并在运行时将结果存储到查找表中。你制作的桌子越大，你使用的内存就越多，但你也会获得更高的精确度。
更独特的方法涉及内存中的浮点表示。其中一个更着名的浮点攻击是fast inverse square root。通过这些概念，一般的想法是你可以为你想要的任何东西做一个近似函数，而不仅仅是反平方根。如果您知道输入范围和误差容差，则可以构建一个精确调整的算法。
如果不在基准测试中说明重要性，我将非常失职！如果您想要应用这些技术中的任何一种来加速您的计划，首先进行基准测试，并确保您正在优化正确的位置！
答案 3 :(得分：0)
是的，当它们是2的幂时，还有用伪掩码进行模运算的可能性，但在这种情况下，我们必须考虑浮动的事实点编号按照IEEE-754标准格式化。
让我们假设我们做同样的操作，但这次，由于数字是实数，两个数字的幂将是1位，后跟无限数量的0第
1000000000000000000000000... * 2^exp
获取掩码，我们执行与整数相同的操作...将1更改为0，并将该数字后面的所有位更改为{ {1}}秒。
1但这是一个全部掩码，因此除了超过模数的位（变为零）之外，永远不会触及尾数。当我们零所有这些时，我们不需要做任何事情，只需将它们移到左边，让它们掉进废纸篓，因为它们被掩盖了。因此，尾数的屏蔽始终是左移（在数字的右边填充更多的数字--- oops，我们没有它们，所以填充零/随机位/ 1 ...）和然后标准化数字（这意味着将数字移位，直到第一个有意义的数字到达第一位并且指数偏差被协调调整）
让我们看一个例子：我们有0111111111111111111111111... * 2^exp =
1111111111111111111111111... * 2^(exp-1)
(THIS NUMBER IS (1.0 - FLT_EPSILON) * 2^(exp_of_module - 1))
作为模数，2.0（M_PI）作为要屏蔽的数字：
3.141592... 3.141592... = 40 49 0f db
= 0100 0000 0100 1001 0000 1111 1101 1011...
; which represents
= 0 (positive)
100 0000 0 (exponent biased 128 ==> 1)
(1.)100 1001 0000 1111 1101 1011...
^ Allways 1 so IEEE-754 doesn't include, we have to do, to operate.
11.00 1001 0000 1111 1101 1011...
01.11 1111 1111 1111 1111 1111...
MASKED == 01.00 1001 0000 1111 1101 1011...
/ // //// //// //// //// ////,-- introduced to complete format
RENORMALIZED == 1.001 0010 0001 1111 1011 011X...
; result = 0 (positive)
= 011 1111 1 (new exponent after norm. 127 ==> 0)
1.001 0010 0001 1111 1011 011X
= 0011 1111 1001 0010 0001 1111 1011 011X
符合预期。
如你所见，我们必须检查有偏差的指数之间的差异，并将尾数移动到该差异所指示的多个位置。同时，我们需要将该差值减去偏差指数（并检查下溢或次正常情况）并重新归一化数字（左移尾数，并递减指数，直到尾数中的第一个转到隐藏比特位置）
我假设模数的偏差指数低于要操作的数字的指数，就像在另一种情况下，掩码是全1，并且数字不受掩码的影响。