C编程中的不动点算法

时间:2012-04-08 23:48:44

标签: c fixed-point

我正在尝试创建一个高精度存储股票价格的应用程序。目前我正在使用双倍这样做。为了节省内存,我可以使用任何其他数据类型吗?我知道这与定点运算有关,但我无法弄明白。

4 个答案:

答案 0 :(得分:51)

定点运算背后的想法是存储值乘以一定量,使用所有微积分的相乘值,并在需要结果时将其除以相同的量。此技术的目的是使用整数算术(int,long ...),同时能够表示分数。

在C中执行此操作的通常且最有效的方法是使用位移位运算符(<<>>)。对于ALU来说,移位是一种非常简单和快速的操作,并且这样做具有在每个移位上乘以(<<<")并将(>>)整数值除以2的特性(此外,可以进行许多移位)与单一的价格完全相同)。当然,缺点是乘数必须是2的幂(这通常不是问题,因为我们并不真正关心那个精确的乘数值)。

现在让我们说我们要使用32位整数来存储我们的值。我们必须选择2乘数的幂。让我们将蛋糕分成两部分,所以说65536(这是最常见的情况,但你可以根据你的精确度需要使用2的任何幂)。这是2 16 ,这里的16意味着我们将使用16个最低有效位(LSB)作为小数部分。其余部分(32 - 16 = 16)用于最高有效位(MSB),即整数部分。

     integer (MSB)    fraction (LSB)
           v                 v
    0000000000000000.0000000000000000

让我们把它放在代码中:

#define SHIFT_AMOUNT 16 // 2^16 = 65536
#define SHIFT_MASK ((1 << SHIFT_AMOUNT) - 1) // 65535 (all LSB set, all MSB clear)

int price = 500 << SHIFT_AMOUNT;

这是您必须存储的值(结构,数据库,等等)。请注意,即使现​​在大多数情况下,int也不一定是C中的32位。也没有进一步声明,它默认签名。您可以将unsigned添加到声明中以确保。更好的是,如果你的代码高度依赖于整数位大小,你可以使用uint32_t或uint_least32_t(在stdint.h中声明)(你可能会介绍一些关于它的黑客攻击)。如果有疑问,请使用typedef作为定点类型,并且您更安全。

如果要对此值进行微积分,可以使用4个基本运算符:+, - ,*和/。您必须记住,在添加和减去值(+和 - )时,还必须移动该值。我们假设我们要为我们的500价格增加10:

price += 10 << SHIFT_AMOUNT;

但是对于乘法和除法(*和/),乘数/除数不得移位。我们想说我们要乘以3:

price *= 3;

现在让我们通过将价格除以4来使事情变得更有趣,这样我们就可以弥补一个非零的小数部分:

price /= 4; // now our price is ((500 + 10) * 3) / 4 = 382.5

这完全是关于规则的。如果您想在任何时候检索实际价格,您必须右移:

printf("price integer is %d\n", price >> SHIFT_AMOUNT);

如果您需要小数部分,则必须将其掩盖:

printf ("price fraction is %d\n", price & SHIFT_MASK);

当然,这个值不是我们可以称之为小数的分数,实际上它是[0 - 65535]范围内的整数。但它完全映射小数部分范围[0 - 0.9999 ...]。换句话说,映射看起来像:0 =&gt; 0,32768 =&gt; 0.5,65535 =&gt; 0.9999 ...

将其视为小数部分的简单方法是在此时使用C内置浮点运算:

printf("price fraction in decimal is %f\n", ((double)(price & SHIFT_MASK) / (1 << SHIFT_AMOUNT)));

但如果你没有FPU支持(硬件或软件),你可以使用这样的新技能来获得完整的价格:

printf("price is roughly %d.%lld\n", price >> SHIFT_AMOUNT, (long long)(price & SHIFT_MASK) * 100000 / (1 << SHIFT_AMOUNT));

表达式中0的数量大致是小数点后所需的位数。考虑到你的分数精度,不要过高估计0的数量(这里没有真正的陷阱,这很明显)。不要使用simple long,因为sizeof(long)可以等于sizeof(int)。如果int是32位,则使用 long long ,因为 long long 保证最小为64位(或者使用int64_t,int_least64_t等,在stdint.h中声明)。换句话说,使用两倍于定点类型的类型,这是公平的。最后,如果您无法访问&gt; = 64位类型,可能需要时间来模拟它们,至少对于您的输出而言。

这些是定点算术背后的基本思想。

小心负值。它有时会变得棘手,尤其是在显示最终价值的时候。此外,C是关于有符号整数的实现定义(即使现在这个问题的平台非常罕见)。您应该始终在您的环境中进行最少的测试,以确保一切按预期进行。如果没有,如果你知道你做了什么,你就可以解决它(我不会在这方面发展,但这与算术移位与逻辑移位和2的补码表示有关)。但是,对于无符号整数,无论​​你做什么都是最安全的,因为无论如何行为都是明确定义的。

另请注意,如果32位整数不能表示大于2 32 - 1的值,则使用2 16 的定点运算将范围限制为2 16 - 1! (并用签名整数将所有这一切除以2,在我们的例子中,我们的可用范围为2 15 - 1)。然后目标是选择适合该情况的SHIFT_AMOUNT。这是整数部分幅度和分数部分精度之间的权衡。

现在有了真正的警告:这种技术绝对不适合精确度最高的领域(金融,科学,军事......)。通常浮点(浮点/双点)通常也不够精确,即使它们具有比定点整体更好的属性。无论值如何,定点具有相同的精度(在某些情况下这可能是一个优势),其中浮点精度与值幅度成反比(即,幅度越低,得到的精度越高......好吧,这个比这更复杂,但你明白了。)浮点数的幅度也大于等价(位数)整数(定点与否),以及高值精度损失的成本(甚至可以达到增加1甚至更高的精度的点数)更大的值根本不起作用,这是整数不可能发生的事情。)

如果您在那些明智的领域工作,那么您最好使用专用于任意精确度的库(请查看gmplib,它是免费的)。在计算科学中,实质上,获得精度大约是用于存储值的位数。你想要高精度吗?使用位。这就是全部。

答案 1 :(得分:5)

我为你看到了两个选择。如果您在金融服务行业工作,那么您的代码可能会遵循标准,以确保精确性和准确性,因此无论内存成本如何,您都必须遵循这些标准。我知道该业务通常资金充足,因此支付更多内存应该不是问题。 :)

如果这是供个人使用,那么为了获得最大精度,我建议您在存储之前使用整数并将所有价格乘以固定因子。例如,如果你想要精确到便士(可能不够好)的东西,将所有价格乘以100,这样你的单位实际上是美分而不是美元并从那里开始。如果你想要更高的精度,请乘以更多。例如,准确到百分之一分(我听说过的标准是常用的),将价格乘以10000(100 * 100)。

现在有了32位整数,乘以10000几乎没有留下大量美元的空间。实际的32位限制为20亿意味着只能表达高达20000美元的价格:2000000000/10000 = 20000.如果你将20000乘以某种东西会变得更糟,因为可能没有空间来保存结果。因此,我建议使用64位整数(long long)。即使你将所有价格乘以10000,仍然有足够的空间来保持较大的价值,即使在乘法中也是如此。

定点技巧是每当你进行计算时,你需要记住每个值实际上是一个基础值乘以常量。在加或减之前,需要将值乘以较小的常量以匹配具有较大常量的值。在你乘以之后,你需要除以某个东西以使结果回到乘以所需的常数。如果你使用2的非幂作为常数,你将不得不做一个整数除法,这在时间上是昂贵的。许多人使用两个幂作为常数,因此他们可以转而不是分裂。

如果这一切看起来很复杂,那就是。我认为最简单的选择是使用双打并在需要时购买更多内存。它们具有53位精度,大约为9千万亿,或几乎16位十进制数。是的,当你与数十亿人合作时,你仍然可能会失去便士,但如果你关心这一点,那么你就不是正确的亿万富翁。 :)

答案 2 :(得分:1)

@Alex给出了一个奇妙的答案here。但是,我想对他的工作进行一些改进,例如,通过演示如何将模拟浮点数(使用整数像浮点数一样)四舍五入到任何所需的小数位。我在下面的代码中演示了这一点。但是,我走得更远,最终写了整个代码教程来教自己定点数学。在这里:

fixed_point_math教程
-类似教程的练习代码,以学习如何进行定点数学运算,仅使用整数进行类似“浮点”的手动打印,   类似于“浮点数”的整数舍入,以及大整数的分数定点数学。

如果您真的想学习定点数学,我认为这是值得仔细研究的有价值的代码,但是我花了整个周末的时间来编写,因此预计可能要花几个小时才能彻底完成它所有。 四舍五入的基础知识可以在顶部找到,并且只需几分钟就可以了解。

GitHub上的完整代码:https://github.com/ElectricRCAircraftGuy/fixed_point_math

或者,在下面(被截断,因为堆栈溢出不允许那么多字符):

/*
fixed_point_math tutorial
- A tutorial-like practice code to learn how to do fixed-point math, manual "float"-like prints using integers only,
  "float"-like integer rounding, and fractional fixed-point math on large integers. 

By Gabriel Staples
www.ElectricRCAircraftGuy.com
- email available via the Contact Me link at the top of my website.
Started: 22 Dec. 2018 
Updated: 25 Dec. 2018 

References:
- https://stackoverflow.com/questions/10067510/fixed-point-arithmetic-in-c-programming

Commands to Compile & Run:
As a C program (the file must NOT have a C++ file extension or it will be automatically compiled as C++, so we will
make a copy of it and change the file extension to .c first):
See here: https://stackoverflow.com/a/3206195/4561887. 
    cp fixed_point_math.cpp fixed_point_math_copy.c && gcc -Wall -std=c99 -o ./bin/fixed_point_math_c fixed_point_math_copy.c && ./bin/fixed_point_math_c
As a C++ program:
    g++ -Wall -o ./bin/fixed_point_math_cpp fixed_point_math.cpp && ./bin/fixed_point_math_cpp

*/

#include <stdbool.h>
#include <stdio.h>
#include <stdint.h>

// Define our fixed point type.
typedef uint32_t fixed_point_t;

#define BITS_PER_BYTE 8

#define FRACTION_BITS 16 // 1 << 16 = 2^16 = 65536
#define FRACTION_DIVISOR (1 << FRACTION_BITS)
#define FRACTION_MASK (FRACTION_DIVISOR - 1) // 65535 (all LSB set, all MSB clear)

// // Conversions [NEVERMIND, LET'S DO THIS MANUALLY INSTEAD OF USING THESE MACROS TO HELP ENGRAIN IT IN US BETTER]:
// #define INT_2_FIXED_PT_NUM(num)     (num << FRACTION_BITS)      // Regular integer number to fixed point number
// #define FIXED_PT_NUM_2_INT(fp_num)  (fp_num >> FRACTION_BITS)   // Fixed point number back to regular integer number

// Private function prototypes:
static void print_if_error_introduced(uint8_t num_digits_after_decimal);

int main(int argc, char * argv[])
{
    printf("Begin.\n");

    // We know how many bits we will use for the fraction, but how many bits are remaining for the whole number, 
    // and what's the whole number's max range? Let's calculate it.
    const uint8_t WHOLE_NUM_BITS = sizeof(fixed_point_t)*BITS_PER_BYTE - FRACTION_BITS;
    const fixed_point_t MAX_WHOLE_NUM = (1 << WHOLE_NUM_BITS) - 1;
    printf("fraction bits = %u.\n", FRACTION_BITS);
    printf("whole number bits = %u.\n", WHOLE_NUM_BITS);
    printf("max whole number = %u.\n\n", MAX_WHOLE_NUM);

    // Create a variable called `price`, and let's do some fixed point math on it.
    const fixed_point_t PRICE_ORIGINAL = 503;
    fixed_point_t price = PRICE_ORIGINAL << FRACTION_BITS;
    price += 10 << FRACTION_BITS;
    price *= 3;
    price /= 7; // now our price is ((503 + 10)*3/7) = 219.857142857.

    printf("price as a true double is %3.9f.\n", ((double)PRICE_ORIGINAL + 10)*3/7);
    printf("price as integer is %u.\n", price >> FRACTION_BITS);
    printf("price fractional part is %u (of %u).\n", price & FRACTION_MASK, FRACTION_DIVISOR);
    printf("price fractional part as decimal is %f (%u/%u).\n", (double)(price & FRACTION_MASK) / FRACTION_DIVISOR,
           price & FRACTION_MASK, FRACTION_DIVISOR);

    // Now, if you don't have float support (neither in hardware via a Floating Point Unit [FPU], nor in software
    // via built-in floating point math libraries as part of your processor's C implementation), then you may have
    // to manually print the whole number and fractional number parts separately as follows. Look for the patterns.
    // Be sure to make note of the following 2 points:
    // - 1) the digits after the decimal are determined by the multiplier: 
    //     0 digits: * 10^0 ==> * 1         <== 0 zeros
    //     1 digit : * 10^1 ==> * 10        <== 1 zero
    //     2 digits: * 10^2 ==> * 100       <== 2 zeros
    //     3 digits: * 10^3 ==> * 1000      <== 3 zeros
    //     4 digits: * 10^4 ==> * 10000     <== 4 zeros
    //     5 digits: * 10^5 ==> * 100000    <== 5 zeros
    // - 2) Be sure to use the proper printf format statement to enforce the proper number of leading zeros in front of
    //   the fractional part of the number. ie: refer to the "%01", "%02", "%03", etc. below.
    // Manual "floats":
    // 0 digits after the decimal
    printf("price (manual float, 0 digits after decimal) is %u.", 
           price >> FRACTION_BITS); print_if_error_introduced(0);
    // 1 digit after the decimal
    printf("price (manual float, 1 digit  after decimal) is %u.%01lu.", 
           price >> FRACTION_BITS, (uint64_t)(price & FRACTION_MASK) * 10 / FRACTION_DIVISOR); 
    print_if_error_introduced(1);
    // 2 digits after decimal
    printf("price (manual float, 2 digits after decimal) is %u.%02lu.", 
           price >> FRACTION_BITS, (uint64_t)(price & FRACTION_MASK) * 100 / FRACTION_DIVISOR); 
    print_if_error_introduced(2);
    // 3 digits after decimal
    printf("price (manual float, 3 digits after decimal) is %u.%03lu.", 
           price >> FRACTION_BITS, (uint64_t)(price & FRACTION_MASK) * 1000 / FRACTION_DIVISOR); 
    print_if_error_introduced(3);
    // 4 digits after decimal
    printf("price (manual float, 4 digits after decimal) is %u.%04lu.", 
           price >> FRACTION_BITS, (uint64_t)(price & FRACTION_MASK) * 10000 / FRACTION_DIVISOR); 
    print_if_error_introduced(4);
    // 5 digits after decimal
    printf("price (manual float, 5 digits after decimal) is %u.%05lu.", 
           price >> FRACTION_BITS, (uint64_t)(price & FRACTION_MASK) * 100000 / FRACTION_DIVISOR); 
    print_if_error_introduced(5);
    // 6 digits after decimal
    printf("price (manual float, 6 digits after decimal) is %u.%06lu.", 
           price >> FRACTION_BITS, (uint64_t)(price & FRACTION_MASK) * 1000000 / FRACTION_DIVISOR); 
    print_if_error_introduced(6);
    printf("\n");


    // Manual "floats" ***with rounding now***:
    // - To do rounding with integers, the concept is best understood by examples: 
    // BASE 10 CONCEPT:
    // 1. To round to the nearest whole number: 
    //    Add 1/2 to the number, then let it be truncated since it is an integer. 
    //    Examples:
    //      1.5 + 1/2 = 1.5 + 0.5 = 2.0. Truncate it to 2. Good!
    //      1.99 + 0.5 = 2.49. Truncate it to 2. Good!
    //      1.49 + 0.5 = 1.99. Truncate it to 1. Good!
    // 2. To round to the nearest tenth place:
    //    Multiply by 10 (this is equivalent to doing a single base-10 left-shift), then add 1/2, then let 
    //    it be truncated since it is an integer, then divide by 10 (this is a base-10 right-shift).
    //    Example:
    //      1.57 x 10 + 1/2 = 15.7 + 0.5 = 16.2. Truncate to 16. Divide by 10 --> 1.6. Good.
    // 3. To round to the nearest hundredth place:
    //    Multiply by 100 (base-10 left-shift 2 places), add 1/2, truncate, divide by 100 (base-10 
    //    right-shift 2 places).
    //    Example:
    //      1.579 x 100 + 1/2 = 157.9 + 0.5 = 158.4. Truncate to 158. Divide by 100 --> 1.58. Good.
    //
    // BASE 2 CONCEPT:
    // - We are dealing with fractional numbers stored in base-2 binary bits, however, and we have already 
    //   left-shifted by FRACTION_BITS (num << FRACTION_BITS) when we converted our numbers to fixed-point 
    //   numbers. Therefore, *all we have to do* is add the proper value, and we get the same effect when we 
    //   right-shift by FRACTION_BITS (num >> FRACTION_BITS) in our conversion back from fixed-point to regular
    //   numbers. Here's what that looks like for us:
    // - Note: "addend" = "a number that is added to another".
    //   (see https://www.google.com/search?q=addend&oq=addend&aqs=chrome.0.0l6.1290j0j7&sourceid=chrome&ie=UTF-8).
    // - Rounding to 0 digits means simply rounding to the nearest whole number.
    // Round to:        Addends:
    // 0 digits: add 5/10 * FRACTION_DIVISOR       ==> + FRACTION_DIVISOR/2
    // 1 digits: add 5/100 * FRACTION_DIVISOR      ==> + FRACTION_DIVISOR/20
    // 2 digits: add 5/1000 * FRACTION_DIVISOR     ==> + FRACTION_DIVISOR/200
    // 3 digits: add 5/10000 * FRACTION_DIVISOR    ==> + FRACTION_DIVISOR/2000
    // 4 digits: add 5/100000 * FRACTION_DIVISOR   ==> + FRACTION_DIVISOR/20000
    // 5 digits: add 5/1000000 * FRACTION_DIVISOR  ==> + FRACTION_DIVISOR/200000
    // 6 digits: add 5/10000000 * FRACTION_DIVISOR ==> + FRACTION_DIVISOR/2000000
    // etc.

    printf("WITH MANUAL INTEGER-BASED ROUNDING:\n");

    // Calculate addends used for rounding (see definition of "addend" above).
    fixed_point_t addend0 = FRACTION_DIVISOR/2;
    fixed_point_t addend1 = FRACTION_DIVISOR/20;
    fixed_point_t addend2 = FRACTION_DIVISOR/200;
    fixed_point_t addend3 = FRACTION_DIVISOR/2000;
    fixed_point_t addend4 = FRACTION_DIVISOR/20000;
    fixed_point_t addend5 = FRACTION_DIVISOR/200000;

    // Print addends used for rounding.
    printf("addend0 = %u.\n", addend0);
    printf("addend1 = %u.\n", addend1);
    printf("addend2 = %u.\n", addend2);
    printf("addend3 = %u.\n", addend3);
    printf("addend4 = %u.\n", addend4);
    printf("addend5 = %u.\n", addend5);

    // Calculate rounded prices
    fixed_point_t price_rounded0 = price + addend0; // round to 0 decimal digits
    fixed_point_t price_rounded1 = price + addend1; // round to 1 decimal digits
    fixed_point_t price_rounded2 = price + addend2; // round to 2 decimal digits
    fixed_point_t price_rounded3 = price + addend3; // round to 3 decimal digits
    fixed_point_t price_rounded4 = price + addend4; // round to 4 decimal digits
    fixed_point_t price_rounded5 = price + addend5; // round to 5 decimal digits

    // Print manually rounded prices of manually-printed fixed point integers as though they were "floats".
    printf("rounded price (manual float, rounded to 0 digits after decimal) is %u.\n", 
           price_rounded0 >> FRACTION_BITS); 
    printf("rounded price (manual float, rounded to 1 digit  after decimal) is %u.%01lu.\n", 
           price_rounded1 >> FRACTION_BITS, (uint64_t)(price_rounded1 & FRACTION_MASK) * 10 / FRACTION_DIVISOR); 
    printf("rounded price (manual float, rounded to 2 digits after decimal) is %u.%02lu.\n", 
           price_rounded2 >> FRACTION_BITS, (uint64_t)(price_rounded2 & FRACTION_MASK) * 100 / FRACTION_DIVISOR); 
    printf("rounded price (manual float, rounded to 3 digits after decimal) is %u.%03lu.\n", 
           price_rounded3 >> FRACTION_BITS, (uint64_t)(price_rounded3 & FRACTION_MASK) * 1000 / FRACTION_DIVISOR); 
    printf("rounded price (manual float, rounded to 4 digits after decimal) is %u.%04lu.\n", 
           price_rounded4 >> FRACTION_BITS, (uint64_t)(price_rounded4 & FRACTION_MASK) * 10000 / FRACTION_DIVISOR); 
    printf("rounded price (manual float, rounded to 5 digits after decimal) is %u.%05lu.\n", 
           price_rounded5 >> FRACTION_BITS, (uint64_t)(price_rounded5 & FRACTION_MASK) * 100000 / FRACTION_DIVISOR); 


    // =================================================================================================================

    printf("\nRELATED CONCEPT: DOING LARGE-INTEGER MATH WITH SMALL INTEGER TYPES:\n");

    // RELATED CONCEPTS:
    // Now let's practice handling (doing math on) large integers (ie: large relative to their integer type),
    // withOUT resorting to using larger integer types (because they may not exist for our target processor), 
    // and withOUT using floating point math, since that might also either not exist for our processor, or be too
    // slow or program-space-intensive for our application.
    // - These concepts are especially useful when you hit the limits of your architecture's integer types: ex: 
    //   if you have a uint64_t nanosecond timestamp that is really large, and you need to multiply it by a fraction
    //   to convert it, but you don't have uint128_t types available to you to multiply by the numerator before 
    //   dividing by the denominator. What do you do?
    // - We can use fixed-point math to achieve desired results. Let's look at various approaches.
    // - Let's say my goal is to multiply a number by a fraction < 1 withOUT it ever growing into a larger type.
    // - Essentially we want to multiply some really large number (near its range limit for its integer type)
    //   by some_number/some_larger_number (ie: a fraction < 1). The problem is that if we multiply by the numerator
    //   first, it will overflow, and if we divide by the denominator first we will lose resolution via bits 
    //   right-shifting out.
    // Here are various examples and approaches.

    // -----------------------------------------------------
    // EXAMPLE 1
    // Goal: Use only 16-bit values & math to find 65401 * 16/127.
    // Result: Great! All 3 approaches work, with the 3rd being the best. To learn the techniques required for the 
    // absolute best approach of all, take a look at the 8th approach in Example 2 below.
    // -----------------------------------------------------
    uint16_t num16 = 65401; // 1111 1111 0111 1001 
    uint16_t times = 16;
    uint16_t divide = 127;

    printf("\nEXAMPLE 1\n");

    // Find the true answer.
    // First, let's cheat to know the right answer by letting it grow into a larger type. 
    // Multiply *first* (before doing the divide) to avoid losing resolution.
    printf("%u * %u/%u = %u. <== true answer\n", num16, times, divide, (uint32_t)num16*times/divide);

    // 1st approach: just divide first to prevent overflow, and lose precision right from the start.
    uint16_t num16_result = num16/divide * times;
    printf("1st approach (divide then multiply):\n");
    printf("  num16_result = %u. <== Loses bits that right-shift out during the initial divide.\n", num16_result);

    // 2nd approach: split the 16-bit number into 2 8-bit numbers stored in 16-bit numbers, 
    // placing all 8 bits of each sub-number to the ***far right***, with 8 bits on the left to grow
    // into when multiplying. Then, multiply and divide each part separately. 
    // - The problem, however, is that you'll lose meaningful resolution on the upper-8-bit number when you 
    //   do the division, since there's no bits to the right for the right-shifted bits during division to 
    //   be retained in.
    // Re-sum both sub-numbers at the end to get the final result. 
    // - NOTE THAT 257 IS THE HIGHEST *TIMES* VALUE I CAN USE SINCE 2^16/0b0000,0000,1111,1111 = 65536/255 = 257.00392.
    //   Therefore, any *times* value larger than this will cause overflow.
    uint16_t num16_upper8 = num16 >> 8; // 1111 1111
    uint16_t num16_lower8 = num16 & 0xFF; // 0111 1001
    num16_upper8 *= times;
    num16_lower8 *= times;
    num16_upper8 /= divide;
    num16_lower8 /= divide;
    num16_result = (num16_upper8 << 8) + num16_lower8;
    printf("2nd approach (split into 2 8-bit sub-numbers with bits at far right):\n");
    printf("  num16_result = %u. <== Loses bits that right-shift out during the divide.\n", num16_result);

    // 3rd approach: split the 16-bit number into 2 8-bit numbers stored in 16-bit numbers, 
    // placing all 8 bits of each sub-number ***in the center***, with 4 bits on the left to grow when 
    // multiplying and 4 bits on the right to not lose as many bits when dividing. 
    // This will help stop the loss of resolution when we divide, at the cost of overflowing more easily when we 
    // multiply.
    // - NOTE THAT 16 IS THE HIGHEST *TIMES* VALUE I CAN USE SINCE 2^16/0b0000,1111,1111,0000 = 65536/4080 = 16.0627.
    //   Therefore, any *times* value larger than this will cause overflow.
    num16_upper8 = (num16 >> 4) & 0x0FF0;
    num16_lower8 = (num16 << 4) & 0x0FF0;
    num16_upper8 *= times;
    num16_lower8 *= times;
    num16_upper8 /= divide;
    num16_lower8 /= divide;
    num16_result = (num16_upper8 << 4) + (num16_lower8 >> 4);
    printf("3rd approach (split into 2 8-bit sub-numbers with bits centered):\n");
    printf("  num16_result = %u. <== Perfect! Retains the bits that right-shift during the divide.\n", num16_result);

    // -----------------------------------------------------
    // EXAMPLE 2
    // Goal: Use only 16-bit values & math to find 65401 * 99/127.
    // Result: Many approaches work, so long as enough bits exist to the left to not allow overflow during the 
    // multiply. The best approach is the 8th one, however, which 1) right-shifts the minimum possible before the
    // multiply, in order to retain as much resolution as possible, and 2) does integer rounding during the divide
    // in order to be as accurate as possible. This is the best approach to use.
    // -----------------------------------------------------
    num16 = 65401; // 1111 1111 0111 1001 
    times = 99;
    divide = 127;

    printf("\nEXAMPLE 2\n");

    // Find the true answer by letting it grow into a larger type.
    printf("%u * %u/%u = %u. <== true answer\n", num16, times, divide, (uint32_t)num16*times/divide);

    // 1st approach: just divide first to prevent overflow, and lose precision right from the start.
    num16_result = num16/divide * times;
    printf("1st approach (divide then multiply):\n");
    printf("  num16_result = %u. <== Loses bits that right-shift out during the initial divide.\n", num16_result);

    // 2nd approach: split the 16-bit number into 2 8-bit numbers stored in 16-bit numbers, 
    // placing all 8 bits of each sub-number to the ***far right***, with 8 bits on the left to grow
    // into when multiplying. Then, multiply and divide each part separately. 
    // - The problem, however, is that you'll lose meaningful resolution on the upper-8-bit number when you 
    //   do the division, since there's no bits to the right for the right-shifted bits during division to 
    //   be retained in.
    // Re-sum both sub-numbers at the end to get the final result. 
    // - NOTE THAT 257 IS THE HIGHEST *TIMES* VALUE I CAN USE SINCE 2^16/0b0000,0000,1111,1111 = 65536/255 = 257.00392.
    //   Therefore, any *times* value larger than this will cause overflow.
    num16_upper8 = num16 >> 8; // 1111 1111
    num16_lower8 = num16 & 0xFF; // 0111 1001
    num16_upper8 *= times;
    num16_lower8 *= times;
    num16_upper8 /= divide;
    num16_lower8 /= divide;
    num16_result = (num16_upper8 << 8) + num16_lower8;
    printf("2nd approach (split into 2 8-bit sub-numbers with bits at far right):\n");
    printf("  num16_result = %u. <== Loses bits that right-shift out during the divide.\n", num16_result);

    /////////////////////////////////////////////////////////////////////////////////////////////////
    // TRUNCATED BECAUSE STACK OVERFLOW WON'T ALLOW THIS MANY CHARACTERS.
    // See the rest of the code on github: https://github.com/ElectricRCAircraftGuy/fixed_point_math
    /////////////////////////////////////////////////////////////////////////////////////////////////

    return 0;
} // main

// PRIVATE FUNCTION DEFINITIONS:

/// @brief A function to help identify at what decimal digit error is introduced, based on how many bits you are using
///        to represent the fractional portion of the number in your fixed-point number system.
/// @details    Note: this function relies on an internal static bool to keep track of if it has already
///             identified at what decimal digit error is introduced, so once it prints this fact once, it will never 
///             print again. This is by design just to simplify usage in this demo.
/// @param[in]  num_digits_after_decimal    The number of decimal digits we are printing after the decimal 
///             (0, 1, 2, 3, etc)
/// @return     None
static void print_if_error_introduced(uint8_t num_digits_after_decimal)
{
    static bool already_found = false;

    // Array of power base 10 values, where the value = 10^index:
    const uint32_t POW_BASE_10[] = 
    {
        1, // index 0 (10^0)
        10, 
        100, 
        1000, 
        10000, 
        100000,
        1000000,
        10000000,
        100000000,
        1000000000, // index 9 (10^9); 1 Billion: the max power of 10 that can be stored in a uint32_t
    };

    if (already_found == true)
    {
        goto done;
    }

    if (POW_BASE_10[num_digits_after_decimal] > FRACTION_DIVISOR)
    {
        already_found = true;
        printf(" <== Fixed-point math decimal error first\n"
               "    starts to get introduced here since the fixed point resolution (1/%u) now has lower resolution\n"
               "    than the base-10 resolution (which is 1/%u) at this decimal place. Decimal error may not show\n"
               "    up at this decimal location, per say, but definitely will for all decimal places hereafter.", 
               FRACTION_DIVISOR, POW_BASE_10[num_digits_after_decimal]);
    }

done:
    printf("\n");
}

输出:

  

gabriel $ cp fixed_point_math.cpp fixed_point_math_copy.c && gcc -Wall -std = c99 -o ./bin/fixed_point_math_c> fixed_point_math_copy.c && ./bin/fixed_point_math_c
  开始。
  分数位=16。
  整数位=16。
  最大整数=65535。

     

价格为真正的两倍是219.857142857。
  价格(整数)是219。
  价格小数部分是56173(总计65536)。
  价格小数部分(如小数)为0.857132(56173/65536)。
  价格(手动浮动,小数点后0位)是219。
  价格(手动浮动,小数点后一位)为219.8。
  价格(手动浮动,小数点后两位)为219.85。
  价格(手动浮动,小数点后3位)为219.857。
  价格(手动浮动,小数点后4位)为219.8571。
  价格(手动浮动,小数点后5位)为219.85713。 <==先定点数学十进制错误
      由于定点分辨率(1/65536)现在具有较低的分辨率,因此开始在这里介绍
      比此小数点后的以10为底的分辨率(即1/100000)高。十进制错误可能不会显示
      可以说在这个小数点后的位置,但绝对会在此后所有小数点后的位置。
  价格(手动浮动,小数点后6位)为219.857131。

     

基于手动整数的环绕:
  addend0 =32768。
  addend1 =3276。
  addend2 =327。
  addend3 =32。
  addend4 =3。
  addend5 =0。
  舍入后的价格(手动浮动,小数点后四舍五入到0位)是
。   舍入后的价格(手动浮动,小数点后四舍五入为1位)是219.9。
  舍入后的价格(手动浮动,四舍五入到小数点后两位)是219.86。
  舍入后的价格(手动浮动,小数点后四舍五入到3位数)为219.857。
  舍入后的价格(手动浮动,四舍五入到小数点后4位)为219.8571。
  舍入后的价格(手动浮动,小数点后四舍五入到5位数字)为219.85713。

     

相关概念:使用小整数类型进行大整数运算:

     

示例1
  65401 * 16/127 =8239。<==正确答案
  第一种方法(相除然后相乘):
    num16_result =8224。<==丢失在初始除法期间右移的位。
  第二种方法(分成2个8位子数字,最右边的位):
    num16_result =8207。<==丢失除法期间右移的位。
  第三种方法(将2个8位子数字以位居中):
    num16_result =8239。<==完美!保留除法期间右移的位。

     

示例2
  65401 * 99/127 =50981。<==真实答案
  第一种方法(相除然后相乘):
    num16_result =50886。<==丢失在初始除法期间右移的位。
  第二种方法(分成2个8位子数字,最右边的位):
    num16_result =50782。<==丢失除法期间右移的位。
  第三种方法(将2个8位子数字以位居中):
    num16_result =1373。<==由于乘法过程中发生溢出而完全错误。
  第4种方法(将4个4位子数字分为中心):
    num16_result =15870。<==由于乘法过程中发生溢出而完全错误。
  第5种方法(将8个2位子编号以位为中心):
    num16_result =50922。<==丢失了一些在除法过程中右移的位。
  第6种方法(将16个1位子数字拆分为左偏的位):
    num16_result =50963。<==丢失除法期间右移的最少位数。
  第7种方法(将16个1位子数字拆分为左偏位):
    num16_result =50963。<== [与第6种方法相同]丢失了除法期间右移的最小位数。
  [全部最佳方法]第8种方法(分割为16个1位子编号,左偏斜,除法时带整数舍入):
    num16_result =50967。<==丢失除法期间右移的最少位数,
    &由于除法过程中的舍入而具有更好的精度。

参考:

答案 3 :(得分:0)

如果您的唯一目的是节省内存,我建议您不要这样做。价格计算中的错误可以累积起来,你就会搞砸它。

如果你真的想要实现类似的东西,你可以只取价格的最小间隔,然后直接使用int和整数运算来操纵你的数字?您只需要在显示时将其转换为浮点数,这样可以让您的生活更轻松。