How to store floating point data in memory

        To understand how floating-point data is stored , Be sure to understand the representation of floating-point numbers in the computer .

(1). According to international IEEE 754 standard Any two-level floating-point number v It can be converted into the following form

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

(-1)^s The sign bit , When s=0,V yes Positive numbers , When s=1,V yes negative .

M be Represents a significant number ,M The value range of is greater than or equal to 1, Less than 2, Because binary data storage will not exceed 2.

2^E be Indicates the index bit .

(2). Let me give you an example

Decimal 5.0, The conversion to binary is 101.0, amount to 1.01*2^2

If converted to the above v The format of , We can draw s=0; M=1.01;E=2,

V=(-1)^0*1.01*2^2

If it is decimal -5.0, According to the above, it can be concluded that binary is -101.0, amount to -1.01*2^2

that S=1;M=1.01;E=2,

V=(-1)^1*-1.01*2^2

(3). according to The international IEEE 754 standard , We can see ：

about 32 A single precision floating-point number of bits , The highest 1 Bits are sign bits S, And then 8 Bits are exponents E, The rest 23 Bits are significant digits M.

about 64 Double precision floating point number of bits , The highest 1 Bits are sign bits S, And then 11 Bits are exponents E, The rest 52 Bits are significant digits M.

(4).IEEE 754 Standard for significant figures M There is a rule , It said M The range of phi is zero 1≤M<2, in other words M It can be written. 1.xxxxxx In the form of , among xxxxxx Represents the fractional part ,IEEE 754 Regulations , Keep it in the computer M when , By default, the first digit of this number is always 1, So it can be discarded , Save only the following fraction .

(5) For index E There are also rules ,E If it is an unsigned integer , hypothesis E by 8 position , Its value range is 0~255; If E by 11 position , Its value range is 0~2047. however , We know , In scientific counting E Yes, you can Now negative , the IEEE 754 in Regulations , In memory E The true value of must be added with an intermediate number ,

about 8 Bit E, This middle number yes 127; about 11 Bit E, In the middle The number is 1023.

give an example ：2^10 Of E yes 10, So save it as 32 When floating-point numbers are in place , must 10+127=137, namely 10001001.

(6). When index E There are also the following three situations when fetching from memory

1. When index E Incomplete 0 Or not all 1 when

When converting an integer to a floating-point type, the exponent E The real value of is ：E Calculated value -127( perhaps 1023)

give an example :0.5 The binary form of is 0.1, Since it is stipulated that the positive part must be 1, Move the decimal point right 1 position , Then for 1.0*2^(-1), Index E by -1+127, Expressed as 01111110, And the mantissa 1.0 Remove the integer part and make it 0, Make up the remaining 0~23 position ：

0 01111110 00000000000000000000000

2. When index E All for 0 when

Index E=1-127 ( perhaps 1-1023) That's the true value , Significant figures M No more first 1, It's reduced to 0.xxxxxx Decimals of , The real value obtained is close to 0.

(3) Index E For all 1 when

If the significant number M All for 0, True values represent ± infinity （ It depends on the sign bit s).

#include<stdio.h>
int main()
{
	int li = 9;
	float* flag = (float*)&li;
	printf("n The value of is :%d\n", li);      //9
	printf("flag The value of is :%f\n", *flag);//0.000000
	// First, reshape 9 Convert to floating point , obtain 0 00000000 00000000000000000001001 Split ,
	// You can get the first sign bit S=0, The next eight digit index E=00000000, The remaining 23 Significant digits of bits M=000 0000 0000 0000 0000 1001
	//V=(-1)^0*0.00000000000000000001001×2^(-126)=1.001×2^(-146)
	//S=0,E=00000000 Namely -126,M=000 0000 0000 0000 0000 1001
	//V Is a close to 0 Positive number of , So decimal decimal means ：0.00000000
	*flag = 9.0;
	printf("n The value of is :%d\n", li);      //1091567616
	//V=(-1)^0*1.001*2^3
	//S=0,E=3+127,M=1.001
	//0 10000010 001 0000 0000 0000 0000 0000
	//V=1091567616  This 32 The binary number of bits , Restore to decimal , Namely  1091567616
	printf("flag The value of is :%f\n", *flag);//9.000000
	return 0;
}