What is binary
Every 2 into a counting rule.
Case:
. 1 public class Demo01 { 2 . 3 public static void main (String [] args) { . 4 / ** . 5 * first saw binary . 6 * / . 7 int I = 50; // 110010 . 8 // in println (i ) output time, Java uses . 9 // the API (method) to convert binary to decimal string 10 System.out.println (I); // "50" . 11 // the Java can be provided a method toBinaryString 12 // see the binary data stored in the memory 13 is System.out.println ( 14 Integer.toBinaryString(i)); 15 for(i=0; i<=50; i++) { 16 System.out.println( 17 Integer.toBinaryString(i)); 18 } 19 20 } 21 }
Computer and binary relations
Hex
Hex for short (abbreviated) binary because binary writing lengthy \ trouble \ error-prone, because hexadecimal is base the whole power of two, so four can be shortened to a binary hex digital.
Abbreviation rules: binary forward from the rear, each 4 bits of a binary number hexadecimal abbreviated.
Case:
. 1 public static void main (String [] args) { 2 / ** . 3 * 2 binary write directly, Java supports after 7 4 * characteristics, binary data can be written to the beginning 0B . 5 * / . 6 int n-= 0b110010 ; . 7 System.out.println (n-); . 8 . 9 // binary writing trouble 10 n-= 0b11010011111101010010111101; . 11 // 16 hexadecimal binary short, convenient to use 12 is 13 is int I = 0x77a65fa8 ; 14 the System.out .println ( 15 Integer.toBinaryString (I)); 16 }
Complement
Computer coding problem solving a negative number (a signed number), which is a core purpose of the fixed number of binary digits, half as a negative fraction.
How will complement binary fixed number of points half as negative?
4-bit binary encoding rules complement an example to explain:
- Computing time, more than four times the automatic overflow give up, always keep the digital binary number is four.
- 1 is a high negative number, as an integer of 0 to a high
- Negative encoded reverse extrapolated by the positive number.
- Negative and positive is complementary symmetry: Gu called complement.
Case 1
1 public static void main(String[] args) { 2 /** 3 * 补码 4 */ 5 int n = -13; 6 System.out.println(Integer.toBinaryString(n)); 7 for(int i=-50; i<0; i++) { 8 System.out.println( 9 Integer.toBinaryString(i)); 10 } 11 }
Case 2: Special value
1 public static void main(String[] args) { 2 int max = Integer.MAX_VALUE; 3 int min = Integer.MIN_VALUE; 4 System.out.println(max); 5 System.out.println(min); 6 System.out.println(Integer.toBinaryString(max)); 7 System.out.println(Integer.toBinaryString(min)); 8 //int n = -1; 9 //int n = 0b11111111111111111111111111111111; 10 int n = 0xffffffff; 11 System.out.println(Integer.toBinaryString(n)); 12 System.out.println (n-); // accordance decimal output Levels: -1 13 is @ overflow results: It may be positive or may be negative! 14 int K = 100 ; 15 System.out.println (K + max); 16 System.out.println (max + K + max); . 17 // above described calculation result is an annular coding complement! 18 }
Complement complementary symmetry:
Calculation principles:
n 00000000 00000000 00000000 00110010 50
~n 11111111 11111111 11111111 11001101 -51
~n+1 11111111 11111111 11111111 11001110 -50
Complementary symmetry test case:
1 int n = 50; 2 int m = ~n + 1; 3 System.out.println(m);//-50 4 System.out.println(Integer.toBinaryString(n)); 5 System.out.println(Integer.toBinaryString(~n)); 6 System.out.println(Integer.toBinaryString(~n+1));
Classic interview questions:
System.out.println(~5);
如上代码的输出结果:( D ) A.5 B.6 C.-5 D.-6
System.out.println(~-5);
如上代码的输出结果:( A ) A.4 B.5 C.6 D.7
Binary arithmetic
calculating signs:
~ 取反
& 与
| 或
>> 右移位
>>> 逻辑右移位
<< 左移位
And & calculation (logical multiplication)
The basic rule: 0 or 0
0 & 0 = 0
0 & 1 = 0
1 & 0 = 0
1 & 1 = 1
Computing time, the number of bits required two numbers thereof, and the corresponding location of the digital calculation operations:
For chestnut:
n = 01110111 01010100 10111111 11110111
m = 00000000 00000000 00000000 11111111 mask
k=n&m 00000000 00000000 00000000 11110111
Significance computed as: above is a mask (Mask) is calculated (resolved calculation), a result which is split n k is the last 8 bits (1 byte)
experiment
int n = 0x7754bff7;
int m = 0xff; //8位掩码(Mask): 1的个数有8个
int k = n&m;
//按照2进制验证结果
System.out.println(Integer.toBinaryString(n));
System.out.println(Integer.toBinaryString(m));
System.out.println(Integer.toBinaryString(k));
>>> Logical right-shift calculation
The binary digits overall shift to the right, the low automatic overflow, high fill 0:
Case:
1 public static void main(String[] args) { 2 /** 3 * 移位计算案例 4 */ 5 int n = 0x7754bff7; 6 int m = n>>>1; 7 int k = n>>>2; 8 int g = n>>>8; 9 int b3 = (n>>>8) & 0xff; 10 System.out.println(Integer.toBinaryString(n)); 11 System.out.println(Integer.toBinaryString(m)); 12 System.out.println(Integer.toBinaryString(k)); 13 System.out.println(Integer.toBinaryString(g)); 14 System.out.println(Integer.toBinaryString(b3)); 15 }
Split byte integer
Case:
. 1 public static void main (String [] args) { 2 / ** . 3 * will be split into an integer int four byte . 4 * Case: an integer long split eight byte . 5 * / . 6 int n-= 0x7745abd7 ; . 7 int B1 = (n->>> 24) & 0xFF ; . 8 int B2 = (n->>> 16) & 0xFF ; . 9 int B3 = (n->>>. 8) & 0xFF ; 10 int B4 & 0xFF = n- ; . 11 System.out.println (Integer.toBinaryString (n-)); 12 is System.out.println (Integer.toBinaryString (B1)); 13 is System.out.println (Integer.toBinaryString (B2)); 14 System.out.println (Integer.toBinaryString (B3)); 15 System.out.println (Integer.toBinaryString (B4)); 16 / ** . 17 * an integer long split eight byte 18 is * / . 19 long L = 0x76ab3fed723e7828L ; 20 is B1 = ( int ) ((L 56 is >>>) & 0xFF ); 21 is B2 = ( int ) ((48 L >>> ) & 0xFF ); 22 is B3 = ( int ) ((L >>> 40) & 0xFF ); 23 is B4 = ( int ) ((L >>> 32) & 0xFF ); 24 int b5 = (int)((l>>>24)& 0xff); 25 int b6 = (int)((l>>>16)& 0xff); 26 int b7 = (int)((l>>>8)& 0xff); 27 int b8 = (int)((l>>>0)& 0xff); 28 //验证... 29 System.out.println(Long.toBinaryString(l)); 30 }
| OR operation (logical addition)
The basic rule: There is a 1
0 | 0 = 0
0 | 1 = 1
1 | 0 = 1
1 | 1 = 1
When calculated, the two digit numbers corresponding to bits aligned or calculated.
For chestnut:
n= 00000000 00000000 00000000 10011101
m= 00000000 00000000 11011111 00000000
k=n|m 00000000 00000000 11011111 10011101
Significance is calculated as: The two numbers n and m are spliced calculated.
verification:
int n = 0x9d;
int m = 0xdf00;
int k = n|m;
//按照2进制输出
<< left shift
The binary number for each bit to the left, the upper overflow, low fill 0
Mathematically calculated shift
Recalling the decimal point moved Computing:
// 89191.
// 891910.
// 8919100.
// 10进制时候, 数字向左移动一次, 数值扩展10倍
// 110010. 50
// 1100100. 100 = 50<<1
// 11001000. 200 = 50<<2
// 2进制时候, 数字向左移动一次, 数值扩大2倍
// 110010. 50
// 11001. 25 = 50>>1
// 1100. 12 = 50>>2 向小方向取整数
Case:
/**
* 移动计算的数学意义
*/
int n = 50;
System.out.println(n);
System.out.println(n<<1);
System.out.println(n<<2);
System.out.println(n<<3);
System.out.println(n>>1);
System.out.println(n>>2);
The difference between >> >>>
>> Known as an arithmetic right shift calculation, the time when the positive (high is 0), the high bit of 0, the time when a negative (upper 1) the high bit of 1, which is a mathematical operation result division, the direction of the small rounding.
>>> Called a logical right-shift calculation, 0s are high regardless of the sign, when a negative result does not conform to the mathematical operation
For chestnut:
n = 11111111 11111111 11111111 11001110 -50
m=n>>1 111111111 11111111 11111111 1100111 -25
g=n>>2 1111111111 11111111 11111111 110011 -13
k=n>>>1 011111111 11111111 11111111 1100111 比最大值小24
>>Close mathematical operation result is: 2 divided by the direction of the smaller rounded.
>>>Simply the number of bits to the right, the result is not considered a mathematical sense, digitally split merge when using pure right.
Case Interview: Alternative multiplication integer multiple of 2, the use of mathematical shift!
n * 32 可以替换为 ( n<<5 )
n / 2 (n>=0) 可以替换为 ( n>>1 )
