byte3

(ascontents) ' Converts an ANSI-encoded string into a Unicode-encoded string Stransi2unicode = "" Len1=lenb (ascontents) If Len1=0 then Exit function For I=1 to Len1 VARCHAR=MIDB (ascontents,i,1) VARASC=ASCB (varchar) If VARASC > 127 Then Stransi2unicode = Stransi2unicode Chr (AscW (MidB (ascontents,i+1,1) varchar)) I=i+1 Else Stransi2unicode = Stransi2unicode Chr (VARASC) End If Next End Function Function Base64Encode (ascontents) ' BASE64 encoding an ANSI-encoded string

, the alignment byte is determined by the largest member in the structure and the value specified by # pragma pack.A = automatic length of the maximum data member in the StructureB = default value; value range: 1, 2, 4, 8This struct alignment byte = min (a, B) struct type{ int a; bool b;};int main(){ type t; cout Running result: size = 8address a = 0x7fff518d64a0address b = 0x7fff518d64a4 Arrangement:A A A AB 0 0 0 6.When the default value is equal to or greater than the length of al

3Lsgroup64 = ""Lsgroupbinary = midb (ascontents, lnposition, 3) Byte1 = ASCB (midb (lsgroupbinary, 1, 1): savebits1 = byte1 and 3Byte2 = ASCB (midb (lsgroupbinary, 2, 1): savebits2 = byte2 and 15Byte3 = ASCB (midb (lsgroupbinary, 3, 1 )) Char1 = midb (sbase_64_characters, (Bytes 1 and 252)/4) + 1, 1)Char2 = midb (sbase_64_characters, (byte2 and 240)/16) or (savebits1 * 16) and HFF) + 1, 1)Char3 = midb (sbase_64_characters, (byte3 and 192)/64) or (sa

= len1-3ElseLen2 = len1End if Lsresult = "" For lnposition = 1 to len2 step 3Lsgroup64 = ""Lsgroupbinary = midb (ascontents, lnposition, 3) Byte1 = ASCB (midb (lsgroupbinary, 1, 1): savebits1 = byte1 and 3Byte2 = ASCB (midb (lsgroupbinary, 2, 1): savebits2 = byte2 and 15Byte3 = ASCB (midb (lsgroupbinary, 3, 1 )) Char1 = midb (sbase_64_characters, (byte1 and 252) \ 4) + 1, 1)Char2 = midb (sbase_64_characters, (byte2 and 240) \ 16) or (savebits1 * 16) and HFF) + 1, 1)Char3 = midb (sbase_

BaseOfLoader. Why is it 0100 h (256) and 4 K? Because this is the base address of the segment, the 0100h must be multiplied by the last 16.Mov es, axPop axMoV byte [bodd], 0MoV BX, 3Mul BXMoV BX, 2Div BXCMP dx, 0JZ label_evenMoV byte [bodd], 1 The above section is the difficulty of the entire function. The fatentry offset of the user's calculated cluster number in the fat table relative to the first fat address. From the book, we can know that every fatentry in fat12 is 12 characters. So it is

four bytes into IEEE754 data} 31 function PackByteToFloat (byte1, byte2, byte3, byte4: byte): Single; 32 var 33 input: array [1 .. 4] of byte; {define an array to store the input four bytes} 34 output: PSingle; 35 begin 36 input [1]: = byte1; 37 input [2]: = byte2; 38 input [3]: = byte3; 39 input [4]: = byte4; 40 output: = Addr (input); {use the address retrieval method for processing} 41 Result: = output

');//unicode>127Character Character7 =NewCharacter (' big ')); Character Character5= ' big '; Character Character6= ' big '; System.out.println ("Character4 = = Character5" + (character4 = = Character5));//falseSystem.out.println ("Character5 = = Character6" + (Character5 = = Character6));//falseSystem.out.println ("character4 = = Character7" + (character4 = = Character7));//false /*** Byte Cache size is 128 to 127 * Autobox Reference comparison is all true*/ for(bytei = Byte.min

Const sbase_64_characters = "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz0123456789+/" Function Base64Encode (ByVal ascontents) Dim lnposition Dim Lsresult Dim Char1 Dim CHAR2 Dim CHAR3 Dim CHAR4 Dim Byte1 Dim Byte2 Dim Byte3 Dim SaveBits1 Dim SaveBits2 Dim lsgroupbinary Dim LSGROUP64 If Len (ascontents) mod 3 > 0 Then ascontents = ascontents String (3-(Len (ascontents) mod 3), "" Lsresult = "" For lnposition = 1 to Len (ascontents) Step 3 LS

1-byte sequenceBecause different computer systems use different byte sequences to store data, the same 4-byte 32-bit integer is stored in memory in a different way. The byte order is divided into small tail byte sequence (Little endian) and large tail byte sequence (big endian), most of the Intel processor uses the small tail byte sequence, the Motorola processor most uses the big tail (large endian) byte order; Small tail is the low byte emissions at the lower end of memory, high byte emissions

function | encryption Const sbase_64_characters = "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz0123456789+/" Function Base64Encode (ByVal ascontents) Dim lnposition Dim Lsresult Dim Char1 Dim CHAR2 Dim CHAR3 Dim CHAR4 Dim Byte1 Dim Byte2 Dim Byte3 Dim SaveBits1 Dim SaveBits2 Dim lsgroupbinary Dim LSGROUP64 If Len (ascontents) mod 3 > 0 Then ascontents = ascontents String (3-(Len (ascontents) mod 3), "" Lsresult = "" For lnposition = 1 to

1 byte orderBecause different computer systems store data in different byte order and a 4-byte 32-bit integer, the memory storage method is different. the byte order is divided into the Little Endian and the Big Endian. Most Intel processors use the Little Endian, and most Motorola processors use the Big Endian; the small tail is the low byte emission at the low end of the memory, and the high byte emission at the high end of the memory. For example, a 4-byte integer with a value of 0x1234567 co

encoding getencoding (filestream Stream) { Return getencoding (stream, encoding. Default ); } /// /// Obtain the encoding method of a text file. /// /// /// /// Public static encoding getencoding (string filename, encoding defaultencoding) { Filestream FS = new filestream (filename, filemode. Open ); Encoding targetencoding = getencoding (FS, defaultencoding ); FS. Close (); Return targetencoding; } /// /// Obtain the encoding method of a text file stream.

getencoding (filestream stream, string defaultencoding){String targetencoding = defaultencoding;If (stream! = NULL stream. length> = 2){// Save the first 4 bytes of the file streamByte byte1 = 0;Byte byte2 = 0;Byte byte3 = 0;Byte byte4 = 0;// Save the current seek locationLong origpos = stream. Seek (0, seekorigin. Begin );Stream. Seek (0, seekorigin. Begin ); Int nbyte = stream. readbyte ();Byte1 = convert. tobyte (nbyte );Byte2 = convert. tobyte (

defaultencoding){Encoding targetencoding = defaultencoding;if (stream! = null stream. Length >= 2){Save the first 4 bytes of a file streambyte byte1 = 0;byte Byte2 = 0;byte byte3 = 0;byte byte4 = 0;Save current seek positionLong Origpos = stream. Seek (0, Seekorigin.begin);Stream. Seek (0, Seekorigin.begin);int nbyte = stream. ReadByte ();Byte1 = Convert.tobyte (nbyte);Byte2 = Convert.tobyte (stream. ReadByte ());if (stream. Length >= 3){

. public static Encoding GetEncoding (String fileName, Encoding defaultencoding) { FileStream fs = new FileStream (FileName, FileMode.Open); Encoding targetencoding = getencoding (FS, defaultencoding); Fs. Close (); return targetencoding; } Gets how a text file stream is encoded. public static Encoding GetEncoding (FileStream stream, Encoding defaultencoding) { Encoding targetencoding = defaultencoding; if (stream!= null stream. Length >= 2) { Save the first 4 bytes of a file stream by

, 0x115b2b19l, 0x020bd8edl, 0xf0605beel, 0x24aa3f05l, 0xd6c1bc06l, 0xc5914ff2l, 0x37faccf1l, 0x69e9f0d5l, 0x9b8273d6l, 0x88d28022l, 0x7ab90321l, 0xae70000cal, 0x5c18e4c9l, 0x4f48173dl, 0xbd23943el, 0xf36e6f75l, 0x0105ec76l, 0x12551f82l, 0xe03e9c81l, 0x34f4f86al, 0xc69f7b69l, 0xd5cf889dl, 0x27a40b9el, 0x79b737bal, 0x8bdcb4b9l, 0x988c474dl, 0x6ae7c44el, 0xbe2da0a5l, 0x4c4623a6l, 0x5f16d052l, 0xad7d5351l, }; # Endif U_int32_t Crc32c (unsigned char * buffer, unsigned int length) { Unsigned int I; Un

should understand the structure of this Code. Next I will introduce the segment descriptor, Global Descriptor Table register, and select the sub-data structure.The Structure of the segment descriptor is as follows:The segment descriptor occupies 8 bytes. Here I just want to remind you that the segment base address occupies byte2, byte3, byte4, and byte7 respectively. We need to use these when initializing the segment descriptor below.The structure of

^ 0xab345678.The Judge function determines whether the two integers satisfy the condition, and if the condition is met, the registration succeeds.Focus is within the judge function.　　The algorithm is concise, so my focus is how to find a qualified num1 and num2. Violent search, search space is too large, we can analyze the algorithm to reduce the scope of the search.V4 form the string String1 need to meet equals String2, and V7 is the machine code, are known.As shown in, *v4++ = num1 ^ (v7–num2

writing those numbers to a file. the two orders are called "little endian" and "Big endian ". The basics "Little endian" means that the low-order byte of the number is stored in memory at the lowest address, and the high-order byte at the highest address. (The little end comes first .) for example, a 4 byte longint Byte3 Byte2 Byte1 Byte0 Will be arranged in memory as follows: Base Address+0 Byte0 Base Address+1 Byte1 Base Address+2

)) Loop Tstream.close () Set Tstream = Nothing If InStr (Filecon, Filecon2) > 0 Then Filecon = Replace (Filecon, Filecon2, "") Set Tstream = CreateObject ("ADODB. Stream ") Tstream.type = 2 Tstream.mode = 3 Tstream.charset = Xset Tstream.open Tstream.position=0 Tstream.writetext Filecon Tstream.savetofile filepath, 2 Tstream.close () Set Tstream = Nothing WScript.Echo "The file has been repaired:" filepath ... " End If End Sub Function Getcharset (XPath) Set Tstream = Creat