About data storage in memory

About the storage of data in memory

This article will explain the knowledge of floating-point numbers stored in memory

First , Let's take a look at the following program :>

While studying , The answer I give is very simple 9 9.0 9 9.0, However , When the program runs ：>

To analyze the above procedures , First of all, we have to understand the storage of floating-point numbers in memory .

For binary floating point numbers , According to international standards IEEE（ Institute of electrical and Electronic Engineering ） 754, Any binary floating point number V Can be expressed in the following form :>

*(-1) ^S * M 2 ^E

among （-1）^S Symbol bit , When s = 0 when ,V Is a positive number , When floating point number s = 1 , V It's a negative number .

M Is the significant bit , M>=1 And M < 2

2 ^E The power represents the exponential bit

for instance :>

For example, decimal floating point numbers 5.5, When we convert it to binary, it is :>

101.1 // Here is .1 The reason is here 1 The weight of is 2 ^-1 Power , That is to say 0.5. And we use it V The format of , Can write like this :>

*(-1)^0 1.011 * 2 ^ 2

namely S = 0; M = 1.011; E = 2;

about -5.5, Use it V The format of , I can express it as

*（-1）^1 (1.011) * 2 ^2;

For the storage of floating-point numbers in memory , There are the following rules

For single precision floating point numbers

For double precision floating point numbers :>

There are also some provisions : because M Always >=1 Less than 2, namely M Always equal to 1 The integer of , therefore , When stored in memory , Only decimal places are saved , example 5.5

It can be expressed as **(-1)^0 1.011 * 2 ^ 2*, M = 1.011. But what we store , It only stores 011, And when we use it , That's what makes 1 Make up for it .

For index E for , The first is a unsigned int Number of types , So its scope is 0 ~ 255（8bit） perhaps 0 ~2047（11bit）

When expressed in scientific counting , It is inevitable that E negative for example Decimal system 0.5 Expressed as a binary number :>

1*10 ^-1 also E The range of is greater than or equal to 0, therefore , Specify storage in memory E The true value of must be added with an intermediate number , The add 127 （8bit） or 1023（11bit） after , Then store it in memory

For example, just now -1., Save as 32 When floating-point numbers are in place , Save as -1 + 127 = 126 namely

01111110

Besides , Index E There are also the following provisions :>

if E The value of is not all 1 or 0, be E The real value of is the value stored in memory minus 127（ or 1023）, And then the significant number M Add the first one before .

For example, just now 0.5 Binary system V Expressed as :>

1.0*2 ^ -1 here E by -1, M by 1.0, Therefore, the data stored in memory is

0 01111110 00000000000000000000000

if E The data stored in is full 0 when , So let's think about it , What a small number it would be ！, namely 2^ -127 Power , therefore , The regulations are as follows :>

When E The data stored in is full 0 when , Its index E The real value of is 1 - 127（ or 1023）, And the significant number M No more one before , But add 0, To represent a close 0 Number of numbers .

When E The data stored in is all data , that E The real value of is 128（8bit）, that 2^E Power is a very large number , therefore ,V Expressed as positive and negative infinity .

Now let's go back to the first example at the beginning :>

#include <stdio.h>

int main() 
{
    
	int n = 9; 
	float* pFloat = (float*)&n;
	printf("n The value of is ：%d\n", n); 
	printf("*pFloat The value of is ：%f\n", *pFloat);

	*pFloat = 9.0; 
	printf("num The value of is ：%d\n", n); 
	printf("*pFloat The value of is ：%f\n", *pFloat); 
	return 0;
}

Take another look at the second part :>

Convert... As a floating point number n Changed to 9.0, that 9.0 The binary representation of :>

1001.0 namely 1.001*2^3 that S = 0; E = 3 + 127 = 130; M = 0.001. namely

0 1000 0010 00100000000000000000000

So in the form of integers , The data is :>

In the form of floating point numbers , Certainly 9.0 了 .