# JAVA Binary Basics

Source: Internet
Author: User
Tags bitwise

1. Decimal Binary Mutual Transfer

2. Binary bit operations

3.JDK built-in conversion

The binary in 4.JAVA

decimal binary Mutual transfer

57 111001

Binary bit operations:Advantages: In specific cases, the calculation is convenient and is widely supported. ① Bitwise AND & (two bits full bit 1, results only 1) 0 and 0=0;0 with 1=0;1 and 0=0; 1 and 1=1; Example: 51 and 500110011---------------special usage of 00000101=00000001=1 bit operations: * zeroing, Take the specified bit in a number ② bitwise OR | (As long as there is a bit 1, the result is 1) 0|0=0; 0|1=1; 1|0=1; 1|1=1:51|5 =00110011---------------00000101=00110111 = 55 or special usage of the operation: it is often used to 1;③ XOR operation ∧ for some locations of a data (two corresponding bits are "XOR", the result is 1, otherwise 0) 0∧0=0; 0∧1=1; 1∧0=1; 1∧1=0:51∧5=00110011---------------00000101=00110110=54 Special Usage: * Make certain bits Flip * with 0 xor or retain original value * Inverse operation (reverse the bitwise of a binary number, turn 0 to 1, 1 to 0) Using the method of exchanging values with 0 XOR, two variables: Two variable exchange values are: 1. With the third variable c=a; A=b; B=c;2. Using addition and subtraction c=a+b; B=c-b; A=c-b;3. Using bit XOR or operation to achieve (highest efficiency) a=a∧b; B =a∧b; A=a∧b; Left shift Operation << (the bits of an operand all left several bits, the left bits discarded, the right side 0) 2<<1= 4 Right Shift operation >> (will be an operand of the bits all right shift several bits, positive left 0, negative left to fill 1, Right discard) Unsigned right-shifted negative numbers represent the original code in the complement of their positive values: a positive number is converted to binary by absolute size called the original code. Anti-code complement: Anti-code plus 1 called the complementJDK built-in conversion:

binary in Java:Usually developed, the conversion and bit operation is not much, because Java processing is high-level in the cross-platform use more, file read and write, data communications and so on. Basic data type: INT data class: 1 bytes = 8bitbyte (8bit,-128~127) short (16bit) int (32bit) long (64bit) float: Single precision 32bit, double precision 64bitbolean:true 1bit,false 1bitchar:unicode character 16-bit corresponding wrapper class: Integer ... Other types that drop the base data type in Java are reference data types, and string is a class, so string is not a base type but a reference type.

Data type converted to byte: Int (8143) 8143 (00000000,00000000,00011111,11001111) =byte[]={-49,31,0,0} first byte (Low end) 8143>>0*8& 0xff= (11001111) =207 or signed-49 two bytes (low end) 8143>>1*8&0xff= (00001111) =31 third byte (Low end) 8143>>2*8&0xff= (00000000) =0 fourth byte (Low end) 8143>>3*8&0xff= (00000000) =0 byte to int why do we have to first &0xff calculate the correct answer? First of all, the binary in Java uses the complement form, not the original code or the inverse code, these 3 concepts to be clear; second, byte is 8 bits, int is 32 bit, when Byte is cast to int, if not done & 0xFF, and byte corresponding value is negative, The upper 3 bytes are interpolated, so there is a possibility of errors in the complement error. For example, a byte type of-1, the second binary (complement) of 11111111 (that is, 0xff), converted to int, the value should also be-1, but after the complement, The resulting binary is 11111111111111111111111111111111 (that is, 0xFFFFFFFF), which is not-1, right? And the 0xFF default is int, so, a byte and 0xFF phase, will first convert that byte to int type operation, so that the result of the high 3 bytes will always be cleared 0, so the result is what we want ~ string into byte data: string S; byte[] bs = S.getbytes (); byte array converted to string: string s = new Stirng (BS); string s = new string (bs,encode);//encode means "gb2312,utf-8" encoding method
`1  Public classConvert {2 3     /**4 * << left shift arithmetic << right shift operation <<< unsigned right shift5      */6 7     //int turn byte[]8      Public Static byte[] Int2bytes (intID) {9         byte[] arr =New byte[4];TenArr[0] = (byte) ((int) (ID >> 0 * 8) & 0xFF); OneARR[1] = (byte) ((int) (ID >> 1 * 8) & 0xFF); AARR[2] = (byte) ((int) (ID >> 2 * 8) & 0xFF); -ARR[3] = (byte) ((int) (ID >> 3 * 8) & 0xFF); -         returnarr; the     } -  -     //byte[] Turn int -      Public Static intBytes2int (byte[] arr) { +         intRs0 = (int) ((Arr[0] & 0xff) << 0 * 8); -         intRs1 = (int) ((Arr[1] & 0xff) << 1 * 8); +         intRS2 = (int) ((Arr[2] & 0xff) << 2 * 8); A         intRS3 = (int) ((Arr[3] & 0xff) << 3 * 8); at         returnRs0 + rs1 + rs2 +Rs3; -     } -  -     //long converted to byte[] -      Public Static byte[] Long4bytes (LongID) { -         byte[] arr =New byte[8]; in          for(inti = 0; i < arr.length; i++) { -Arr[i] = (byte) ((int) (ID >> i * 8) & 0xFF); to         } +         returnarr; -     } the  *     /** \$ * byte[] conversion to long and conversion int, just long is 64bit, and int is 32bitPanax Notoginseng      */ -      Public Static voidMain (string[] args) { the         byte[] arr = Convert.int2bytes (8143); +System.out.println (arr[0] + "\ n" + arr[1] + "\ n" + arr[2] + "\ n" + arr[3]); A         intrs =Convert.bytes2int (arr); the System.out.println (RS); +  -         byte[] arr2 = Convert.long4bytes (20); \$          for(byteb:arr2) { \$ System.out.println (b); -         } -  the         //string and byte array: -String describle = "I am a string";Wuyi         byte[] Barr =describle.getbytes (); theString des =NewString (Barr); - System.out.println (DES); Wu     } -  About}`
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JAVA Binary Basics

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