如何重新present浮点数在内存用C [英] How to represent FLOAT number in memory in C

问题描述

在阅读教程我碰到了如何在内存中重新present浮点数来了。本教程有一个浮点数的例子。

While reading a tutorial I came across how to represent Float number in memory. The tutorial had an example with a floating point number.

   float a=5.2  with below Diagram

谁能告诉这5.2如何转换到二进制和它是如何重新$ P $上图中以上内存psented吗？

Can anyone please tell how this 5.2 is converted in to binary and how it is represented in memory in above the above diagram?

推荐答案

至于有人说，5.2重新psented为符号位，指数和尾数$ P $。你怎么带code 5.2？

As was said, 5.2 is represented as a sign bit, an exponent and a mantissa. How do you encode 5.2?

5很简单：

101. 

剩下的，为0.2 1/5，因此通过5（或 1.00000划分 0.FFFFFFF ... （十六进制）... （十六进制）），你会得到 0.3333333 ... （十六进制）。

The rest, 0.2 is 1/5, so divide 0.FFFFFFF... (hex) by 5 (or 1.00000... (hex)) and you get 0.3333333... (hex).

这应该给你

0.0011001100110011001100110011... 

添加5，你会得到

Add 5, and you get

101.00110011001100110011...         exp 0    (5.2 * 2^0)

现在接班（标准化的话），并相应调整指数

Now shift it (normalize it) and adjust the exponent accordingly

1.010011001100110011001100110011... exp +2   (1.3 * 2^2)

现在，你只需要加127偏置（即 129 = 0b10000001 ）的指数，并将其存储：

Now you only have to add the bias of 127 (i.e. 129 = 0b10000001) to the exponent and store it:

0 10000001 1010 0110 0110 0110 0110 0110 

忘记尾数的顶部1（始终应该是1，除了一些特殊值，所以它不存储），你会得到：

Forget the top 1 of the mantissa (which is always supposed to be 1, except for some special values, so it is not stored), and you get:

01000000 10100110 01100110 01100110

现在，你只需要决定很少或大端。

Now you only have to decide little or big endian.

这不正是它是如何工作的，但或多或​​少像时5.2一个数字转换成二进制会发生什么。

This is not exactly how it works, but that is more or less what happens when a number like 5.2 is converted to binary.

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