如何规范尾数 [英] How to normalize a mantissa

问题描述

我正在尝试将 int 转换为自定义浮点型，其中用户指定为exp和尾数保留的位数，但我没有了解转换的工作原理。我的函数接受一个int值和一个int exp来表示数字（值* 2 ^ exp），即value = 12，exp = 4，返回192。但是我不了解更改这些值的过程。我已经研究了好几天了，并且正在使用IEEE转换器网络应用程序，但是我只是不了解标准化过程是什么。就像我看到它的移动二进制点并调整指数一样，但是我不知道这是什么意思，有人可以举一个例子吗？另外我也不明白指数偏差是什么。我仅有的信息是，您只需在指数中添加一个数字，但我不明白为什么。我一直在Google搜索我可以理解的示例，但这对我来说没有任何意义

I'm trying to convert an int into a custom float, in which the user specifies the amount of bits reserved for the exp and mantissa, but I don't understand how the conversion works. My function takes in an int value and and int exp to represent the number (value * 2^exp) i.e value = 12, exp = 4, returns 192. but I don't understand the process I need to do to change these. I've been looking at this for days and playing with IEEE converter web apps but I just don't understand what the normalization process is. Like I see that its "move the binary point and adjust the exponent" but I have no idea what this means, can anyone give me an example to go off of? Also I don't understand what the exponent bias is. The only info I have is that you just add a number to your exponent but I don't understand why. I've been searching Google for an example I can understand but this just isn't making any sense to me

推荐答案

当我们强制将其尾数的整数部分精确地设为 1 并允许其分数部分为我们想要的时，点号为归一化。

A floating point number is normalized when we force the integer part of its mantissa to be exactly 1 and allow its fraction part to be whatever we like.

例如，如果我们采用数字 13.25 ，即 1101.01 二进制， 1101 是整数部分， 01 是小数部分。

For example, if we were to take the number 13.25, which is 1101.01 in binary, 1101 would be the integer part and 01 would be the fraction part.

我可以将 13.25 表示为 1101.01 *（2 ^ 0） ，但这并未标准化，因为整数部分不是 1 。 但是，如果我们将指数增加1：我们可以将尾数向右移一位。

I could represent 13.25 as 1101.01*(2^0), but this isn't normalized because the integer part is not 1. However, we are allowed to shift the mantissa to the right one digit if we increase the exponent by 1:

  1101.01*(2^0)
= 110.101*(2^1)
= 11.0101*(2^2)
= 1.10101*(2^3)

此表示 1.10101 *（2 ^ 3）是归一化的 13.25 的形式。

This representation 1.10101*(2^3) is the normalized form of 13.25.

也就是说，我们知道已归一化浮点数的格式始终为 1.fffffff *（2 ^ exp）

That said, we know that normalized floating point numbers will always come in the form 1.fffffff * (2^exp)

为了效率起见，我们不必费心在二进制表示形式本身中存储 1 整数部分，我们只是假装它在那里。因此，如果要给您的定制浮点型5个尾数位，我们知道 10100 位实际上代表 1.10100 。

For efficiency's sake, we don't bother storing the 1 integer part in the binary representation itself, we just pretend it's there. So if we were to give your custom-made float type 5 bits for the mantissa, we would know the bits 10100 would actually stand for 1.10100.

以下是标准23位尾数的示例：

Here is an example with the standard 23-bit mantissa:

关于指数偏差，让我们看一下标准的32位 float 格式，该格式分为3部分： 1个符号位，8个指数位和23个尾数位：

As for the exponent bias, let's take a look at the standard 32-bit float format, which is broken into 3 parts: 1 sign bit, 8 exponent bits, and 23 mantissa bits:

s eeeeeeee mmmmmmmmmmmmmmmmmmmmmmm

指数 00000000 和 11111111 具有特殊用途（例如表示 Inf 和 NaN ），因此使用8个指数位，我们可以表示254个不同的指数，例如说 2 ^ 1 到 2 ^ 254 。但是，如果我们想代表 2 ^ -3 怎么办？我们如何获得负指数？

The exponents 00000000 and 11111111 have special purposes (like representing Inf and NaN), so with 8 exponent bits, we could represent 254 different exponents, say 2^1 to 2^254, for example. But what if we want to represent 2^-3? How do we get negative exponents?

该格式通过自动从指数中减去127来解决此问题。因此：

The format fixes this problem by automatically subtracting 127 from the exponent. Therefore:

• 0000 0001 将是 1 -127 = -126


• 0010 1101 将是 45 -127 = -82


• 0111 1111 将是 127-127 = 0


• 1001 0010 将是 136-127 = 9

• 0000 0001 would be 1 -127 = -126
• 0010 1101 would be 45 -127 = -82
• 0111 1111 would be 127-127 = 0
• 1001 0010 would be 136-127 = 9

这会将指数范围从 2 ^ 1 ... 2 ^ 254 更改到 2 ^ -126 ... 2 ^ + 127 ，所以我们可以表示负指数。

This changes the exponent range from 2^1 ... 2^254 to 2^-126 ... 2^+127 so we can represent negative exponents.

这篇关于如何规范尾数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！