Binary representation of negative integers and floating point numbers

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Preface

For this code

public class hello {
    
    public static void main(String[] args) {
    
        int a = -17;
        System.out.println(" Binary output "+Integer.toBinaryString(a));
        int b = 17;
        System.out.println(" Binary output "+Integer.toBinaryString(b));
    }
}

Output result

So I'm curious about the binary representation of negative integers and floating-point numbers .

Because of learning 《 Deep understanding of computer systems 》 This course , In chapter two 《 Expression and processing of information 》 I have come into contact with relevant knowledge , The representation of floating point numbers in computers , And the addition and subtraction of signed and unsigned integers . Therefore, it is hereby supplemented .

One 、 Binary representation of negative integers

1. Original code

The original code is the binary representation of the absolute value of the number . In this code ,-17 The original code of is 17 Binary system 10001,int Type of data takes up four bytes, that is 32 position , But omit the previous 0.

2. Inverse code

Inverse code is to reverse the original code by bits , Such as 1 Turn into 0,0 Turn into 1. In this code ,10001 The inverse of is 11111111111111111111111111101110.

3. Complement code

The complement is about to add one to the inverse . In this code ,11111111111111111111111111101110 Add 1 by 11111111111111111111111111101111. It is the binary expression of negative integer .

Two 、 Binary representation of floating-point numbers

1. The formula ：

According to international standards IEEE 754, Any binary floating point number V It can be expressed in the following form ：
（1） (-1)^s The sign bit , When s=0,V Is a positive number ; When s=1,V It's a negative number .
（2） mantissa M Represents a significant number , It's a binary decimal , Its scope is 1~2-ε perhaps 0 ~ 1-ε.
（3）2^E Indicates the index bit ,E It's a step code .

for example ：
take float=1.25f Convert to binary .
s=0;M=1.25;E=2;
about float type . The first is the sign bit , Next 8 Bit is the order code part , The rest of the 23 Bit is the mantissa part .
1. First, convert the number into binary number .

 First step ：
 Bounded by the decimal point , Divided into integer and decimal parts .
 In this question, it is divided into 1 and 0.25 Two parts .
 The second step ：
 The integer part is divided by two and the decimal part is multiplied by two .
 The whole part is 1, The decimal part is 01.
 The third step ：
 Combine integer and decimal parts . namely 1.01.

2. Convert binary decimals to IEEE Floating point standard format V

3. Index part ： Actual index value 2 Add a （ Exponential offset ）, about 32 For single precision floating-point numbers , The offset value is 127, therefore exponent The value is 127 + 2 = 129 , Binary representation as 10000001.

4. Effective word bit , Also called mantissa , namely 2. Binary in represents the decimal part of the value , namely 111101001, Make it up to 23 position , Or get fraction The position is expressed as ：01000000000000000000000

5. result 0 10000001 01000000000000000000000.

2. Classification of floating point numbers （ Classify according to the order code ）

1： Normalized values

This is the most common , When exp The value of is not all 1 Or not all of them 0 This is always the case . At this moment , The order code field is interpreted as a signed integer in the form of an offset , namely E=e-Bias,e Is an unsigned number ,Bias=（2 Of k Minus one power ） Minus one .（k Is the order code point , Determined by the type of the number , If such float It is also a single precision floating-point type , be k The value of is 8; Double precision floating point k The value of is 16.M The value of is 1+f.

2： Denormalized values

When the order code field is all 0 when , For denormalization . Order code value E=1-Bias, The value of the mantissa M=f, Is the value of the decimal field , Does not contain implied 1.

3： Special values

When the order codes are all 1 when , For this situation .
When the decimal fields are all 0 when , The value obtained is infinite , Positive and negative infinity depends on the sign bit s Value .
When the decimal field is non-zero , The result is called “NAN”, namely “Not a Number”.

3. Rounding of floating point numbers

The rounding of floating-point numbers has a directed even rounding , towards 0 Rounding and rounding up or down , In the face of different situations , They are probably the closest rounding method to the real value .