Floating point memory storage problem

Let's first look at the storage of integers

比如int i = 1,Take a look at its actual memory storage 

Then let's take a look at the storage of floating point numbers

Floating point numbers are not directly stored in memory in binary form like integers,After parsing it according to a standard,stored in memory

For example, a floating point number is19.625

Its binary form is：10011.101

Its memory distribution is not as follows

 Generally, we will convert this floating point number into the following form：

 That is, it is stored in memory in the form of an index.

那么19.625In the form of the above, it is

Or look at its binary first：10011.101

The decimal point four left：1.0011101 * 2 ^4

This is stored in memory,So how exactly is it stored in bits?

先来看一下,A Bitwise Storage Rule for Single and Double Precision Floating Point Numbers

 大致意思就是

 

for this number：1.0011101 * 2 ^4

符号位：0,Because it's a positive number

尾数位：0011101,The number of digits after the decimal point,This before the decimal point1呢,Just give up,哈哈

Here's the code：

        阶码 = 阶数 + 偏移量

                阶数 = 指数,Here the exponent is4,二进制就是100

                To calculate the offset first we must know the following two information

                       1.Number of exponents in single precision 8

                       2.Double precision exponent digits11

                We must know the number of exponents,To calculate the offset,So how to calculate the offset？

                2^e-1 - 1,这里eis the number of digits.

                Then the single-precision offset is：2^7 - 1 = 128 -1 =127,The double-precision offset is：2^11 - 1= 1024 - 1 = 1023

           综上所述：code here = 4 + 127（Here I treat it as a single-precision floating-point number） = 131 ,二进制就是：10000011

所以19.625The actual memory distribution of the form is

OK,Then how do we store it in memory?？

         when considering storage,Still have to pay attention to the low byte stored in the low byte,High-order bits are stored in high-order bytes,还要注意,数据的存储顺序（Low order from left to right）with parsing order（从右往左解析）

        

上代码,Let's take a look at the memory

#include <stdio.h>
#include <stdlib.h>

void main()
{
	float num = 19.625f;

	printf("%p", &num);

	system("pause");

	return 0;
}

 go to memory

 Next I talk about the storage of negative floating point numbers,还是拿19.625来说

定义一个数据类型float num = -19.625

直接上代码,然后看内存

#include <stdio.h>
#include <stdlib.h>

void main()
{
	float num = -19.625f;

	printf("%p", &num);

	system("pause");

	return 0;
}

内存：

how did it becomec1了？

Let's take a look at the memory distribution map

换句话说,Negative floating point numbers are to store the source code directly.

Here's another question,精度缺失问题,比如10.6,this fractional part0.6,We're going to be binary

0.6 * 2 = 1.2

0.2 * 2 = 0.4

0.4 * 2 = 0.8

0.8 * 2 = 1.6

0.6....开始循环1001

对于内存来说,It's not allowed to keep looping like this,所以 ,In limited space accordingly,Only certain data can be stored,The latter will be lost directly,所以,This is also the reason why decimals cause certain errors