舍入以使用int - > float - > int往返转换 [英] Rounding to use for int -> float -> int round trip conversion

问题描述

我为数据库引擎编写一组数字类型转换函数，我担心将大型积分浮点值转换为整数类型的行为更精确。

I'm writing a set of numeric type conversion functions for a database engine, and I'm concerned about the behavior of converting large integral floating-point values to integer types with greater precision.

例如将32位int转换为32位单精度浮点数。浮点的23位有效位数产生大约7个精度的十进制数字，因此转换任何int大于大约7位数将导致精度损失（这是精细和预期）。但是，当你把这样一个float转换为int，你会得到它的二进制表示形式的工件的低位数：

Take for example converting a 32-bit int to a 32-bit single-precision float. The 23-bit significand of the float yields about 7 decimal digits of precision, so converting any int with more than about 7 digits will result in a loss of precision (which is fine and expected). However, when you convert such a float back to an int, you end up with artifacts of its binary representation in the low-order digits:

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{
    int a = 2147483000;
    cout << a << endl;
    float f = (float)a;
    cout << setprecision(10) << f << endl;
    int b = (int)f;
    cout << b << endl;

    return 0;
}

打印：

2147483000
2147483008
2147483008

尾部008超出了浮点的精度，因此在int中保留似乎不受欢迎，因为在数据库应用程序中，用户主要关注十进制表示，尾随0用于表示无效数字。

The trailing 008 is beyond the precision of the float, and therefore seems undesirable to retain in the int, since in a database application, users are primarily concerned with decimal representation, and trailing 0's are used to indicate insignificant digits.

所以我的问题是：有没有任何知名的现有系统，执行十进制有效数字舍入在float - > int（或双 - 长long）转换，并有任何好（注意：我知道一些系统具有十进制浮点类型，例如由

So my questions are: Are there any well-known existing systems that perform decimal significant digit rounding in float -> int (or double -> long long) conversions, and are there any well-known, efficient algorithms for doing so?

http://en.wikipedia.org/wiki/IEEE%5F754-2008 =nofollow> IEEE 754-2008 。然而，他们没有主流硬件支持，并且没有内置C / C ++。我可能想支持他们在路上，但我仍然需要直观地处理二进制浮点。）

(Note: I'm aware that some systems have decimal floating-point types, such as those defined by IEEE 754-2008. However, they don't have mainstream hardware support and aren't built into C/C++. I might want to support them down the road, but I still need to handle binary floats intuitively.)

推荐答案

std :: numeric_limits< float> :: digits10 表示您只能获取6个精确的浮点数字。

std::numeric_limits<float>::digits10 says you only get 6 precise digits for float.

高效的算法为您的语言，处理器和数据分配到 calculate-the-decimal-length-of-an-integer （或）。然后减去 digits10 说的精确数字，以获得要剔除的数字数。使用它作为索引，查找10的幂用作模数。 etc。

Pick an efficient algorithm for your language, processor, and data distribution to calculate-the-decimal-length-of-an-integer (or here). Then subtract the number of digits that digits10 says are precise to get the number of digits to cull. Use that as an index to lookup a power of 10 to use as a modulus. Etc.

一个问题：假设你将一个float转换为一个小数并执行这种舍入或截断。然后将调整后的十进制转换为浮点值，并返回到具有相同舍入/截断方案的十进制值。你得到相同的十进制值吗？希望是的。

One concern: Let's say you convert a float to a decimal and perform this sort of rounding or truncation. Then convert that "adjusted" decimal to a float and back to a decimal with the same rounding/truncation scheme. Do you get the same decimal value? Hopefully yes.

这不是你真正想要的，但可能有趣的阅读：向C ++标准库添加最大有效十进制数字值的提案数字限制

This isn't really what you're looking for but may be interesting reading: A Proposal to add a max significant decimal digits value to the C++ Standard Library Numeric limits

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