Atheros無線芯片AR-6000系列wince 6驅動源代碼(這東西我也沒用過別問我,我是搜別的wince資源搜到的) AR6K SDIO support. Requires firmware 1.1 on SD13 cards. readme: Atheros Communications AR6001 WLAN Driver for SDIO installation Read Me March 26,2007 (based on k14 fw1.1) Windows CE Embedded CE 6.0 driver installation. 1. Unzip the installation file onto your system (called installation directory below) 2. Create an OS design or open an existing OS design in Platform Builder 6.0. a. The OS must support the SD bus driver and have an SD Host Controller driver (add these from Catalog Items). b. Run image size should be set to allow greater than 32MB. 3. a. From the Project menu select Add Existing Subproject... b. select AR6K_DRV.pbxml c. select open This should create a subproject within your OS Design project for the AR6K_DRV driver. 4. Build the solution. 轉自Tony嵌入式,原文地址：http://www.cevx.com/bbs/dispbbs.asp?boardID=4&ID=11762&page=1
標簽： wince Requires firmware Atheros
上傳時間： 2014-11-11
上傳用戶：bibirnovis
RSA算法是第一個能同時用于加密和數字簽名的算法，也易于理解和操作。RSA是被研究得最廣泛的公鑰算法，從提出到現在已近二十年，經歷了各種攻擊的考驗，逐漸為人們接受，普遍認為是目前最優秀的公鑰方案之一。RSA的安全性依賴于大數的因子分解，但并沒有從理論上證明破譯RSA的難度與大數分解難度等價。即RSA的重大缺陷是無法從理論上把握它的保密性能如何，而且密碼學界多數人士傾向于因子分解不是NPC問題。 RSA的缺點主要有：A)產生密鑰很麻煩，受到素數產生技術的限制，因而難以做到一次一密。B)分組長度太大，為保證安全性，n 至少也要 600 bits 以上，使運算代價很高，尤其是速度較慢，較對稱密碼算法慢幾個數量級；且隨著大數分解技術的發展，這個長度還在增加，不利于數據格式的標準化。目前，SET( Secure Electronic Transaction )協議中要求CA采用比特長的密鑰，其他實體使用比特的密鑰
上傳時間： 2014-10-13
上傳用戶：sz_hjbf
Atheros Communications AR6001 WLAN Driver for SDIO installation Read Me March 26,2007 (based on k14 fw1.1) Windows CE Embedded CE 6.0 driver installation. 1. Unzip the installation file onto your system (called installation directory below) 2. Create an OS design or open an existing OS design in Platform Builder 6.0. a. The OS must support the SD bus driver and have an SD Host Controller driver (add these from Catalog Items). b. Run image size should be set to allow greater than 32MB. 3. a. From the Project menu select Add Existing Subproject... b. select AR6K_DRV.pbxml c. select open This should create a subproject within your OS Design project for the AR6K_DRV driver. 4. Build the solution.
標簽： Communications installation Atheros Driver
上傳時間： 2016-10-27
上傳用戶：hebmuljb
AR6001 WLAN Driver for SDIO installation Read Me March 26,2007 (based on k14 fw1.1) Windows CE Embedded CE 6.0 driver installation. 1. Unzip the installation file onto your system (called installation directory below) 2. Create an OS design or open an existing OS design in Platform Builder 6.0. a. The OS must support the SD bus driver and have an SD Host Controller driver (add these from Catalog Items). b. Run image size should be set to allow greater than 32MB. 3. a. From the Project menu select Add Existing Subproject... b. select AR6K_DRV.pbxml c. select open This should create a subproject within your OS Design project for the AR6K_DRV driver. 4. Build the solution.
標簽： installation Windows Driver March
上傳時間： 2014-09-06
上傳用戶：yuzsu
Learn how to: * Tokenize a null-terminated string * Create a search and replace function for strings * Implement subtraction for string objects * Use the vector, deque, and list sequence containers * Use the container adaptors stack, queue, and priority_queue * Use the map, multimap, set, and multiset associative containers * Reverse, rotate, and shuffle a sequence * Create a function object * Use binders, negators, and iterator adapters * Read and write files * Use stream iterators to handle file I/O * Use exceptions to handle I/O errors * Create custom inserters and extractors * Format date, time, and numeric data * Use facets and the localization library * Overload the [ ], ( ), and -> operators * Create an explicit constructor * And much, much more
標簽： null-terminated Tokenize Create string
上傳時間： 2014-01-18
上傳用戶：yph853211
RSA算法是第一個能同時用于加密和數字簽名的算法，也易于理解和操作。 RSA是被研究得最廣泛的公鑰算法，從提出到現在已近二十年，經歷了各種攻擊的考驗，逐漸為人們接受，普遍認為是目前最優秀的公鑰方案之一。RSA的安全性依賴于大數的因子分解，但并沒有從理論上證明破譯RSA的難度與大數分解難度等價。即RSA的重大缺陷是無法從理論上把握它的保密性能如何，而且密碼學界多數人士傾向于因子分解不是NPC問題。RSA的缺點主要有：A)產生密鑰很麻煩，受到素數產生技術的限制，因而難以做到一次一密。B)分組長度太大，為保證安全性，n 至少也要 600 bits以上，使運算代價很高，尤其是速度較慢，較對稱密碼算法慢幾個數量級；且隨著大數分解技術的發展，這個長度還在增加，不利于數據格式的標準化。目前，SET(Secure Electronic Transaction)協議中要求CA采用2048比特長的密鑰，其他實體使用1024比特的密鑰。 這種算法1978年就出現了，它是第一個既能用于數據加密也能用于數字簽名的算法。它易于理解和操作，也很流行。算法的名字以發明者的名字命名：Ron Rivest, AdiShamir 和Leonard Adleman。但RSA的安全性一直未能得到理論上的證明。
上傳時間： 2014-01-20
上傳用戶：蠢蠢66
購物車系統中的名牌產品類,private Integer id private String enName private String cnName private String smallPhoto private String bigPhoto private String description private Set products = new HashSet(0)
標簽： private String Integer enName
上傳時間： 2013-12-22
上傳用戶：璇珠官人
#include <iostream.h> #include <string.h> #include <iomanip.h> #include "Stud.h" Stud::Stud(){} char *Stud::getno() //獲取學號 { return no; } char *Stud::getname() //獲取姓名 { return name; } char *Stud::getsex() //獲取性別 { return sex; } char *Stud::getminzu() //獲取民族 { return minzu; } char *Stud::getaddress() //獲取出生地 { return address; } char *Stud::getbirth() //獲取出生年月 { return birth; } int Stud::gettag() //獲取姓名 { return tag; } void Stud::changeno(char n[]) //設置學號 { strcpy(no,n); } void Stud::changename(char na[]) //設置姓名 { strcpy(name,na); } void Stud::changesex(char s[]) //設置性別 { strcpy(sex,s); } void Stud::changeminzu(char m[]) //設置民族 { strcpy(minzu,m); } void Stud::changeaddress(char a[]) //設置出生地 { strcpy(address,a); } void Stud::changebirth(char b[]) //設置出生年月 { strcpy(birth,b); } void Stud::addstudent(char *rn,char *rna) //增加學生 { strcpy(no,rn); strcpy(name,rna); } void Stud::addstudent(char *rn,char *rna,char *rs,char *rm,char *ra,char *rb) //增加學生 { tag=0; strcpy(no,rn); strcpy(name,rna); strcpy(sex,rs); strcpy(minzu,rm); strcpy(address,ra); strcpy(birth,rb); } void Stud::delstud() //設置刪除標記 { tag=1; } void Stud::disp() //輸出學生信息 { cout<<setw(15)<<no<<setw(10)<<name<<setw(10)<<sex<<setw(10)<<minzu<<setw(10)<<address<<setw(10)<<birth<<endl; } void Stud::display() //輸出學生信息 { cout<<setw(15)<<no<<setw(10)<<name; }
標簽： 學生
上傳時間： 2016-12-29
上傳用戶：767483511
/****************temic*********t5557***********************************/ #include <at892051.h> #include <string.h> #include <intrins.h> #include <stdio.h> #define uchar unsigned char #define uint unsigned int #define ulong unsigned long //STC12C2051AD的SFR定義 sfr WDT_CONTR = 0xe1;//stc2051的看門狗?????? /**********全局常量************/ //寫卡的命令 #define write_command0 0//寫密碼 #define write_command1 1//寫配置字 #define write_command2 2//密碼寫數據 #define write_command3 3//喚醒 #define write_command4 4//停止命令 #define TRUE 1 #define FALSE 0 #define OK 0 #define ERROR 255 //讀卡的時間參數us #define ts_min 250//270*11.0592/12=249//取近似的整數 #define ts_max 304//330*11.0592/12=304 #define t1_min 73//90*11.0592/12=83:-10調整 #define t1_max 156//180*11.0592/12=166 #define t2_min 184//210*11.0592/12=194 #define t2_max 267//300*11.0592/12=276 //***********不采用中斷處理:采用查詢的方法讀卡時關所有中斷****************/ sbit p_U2270B_Standby = P3^5;//p_U2270B_Standby PIN=13 sbit p_U2270B_CFE = P3^3;//p_U2270B_CFE PIN=6 sbit p_U2270B_OutPut = P3^7;//p_U2270B_OutPut PIN=2 sbit wtd_sck = P1^7;//SPI總線 sbit wtd_si = P1^3; sbit wtd_so = P1^2; sbit iic_data = P1^2;//lcd IIC sbit iic_clk = P1^7; sbit led_light = P1^6;//測試綠燈 sbit led_light1 = P1^5;//測試紅燈 sbit led_light_ok = P1^1;//讀卡成功標志 sbit fengmingqi = P1^5; /***********全局變量************************************/ uchar data Nkey_a[4] = {0xA0, 0xA1, 0xA2, 0xA3};//初始密碼 //uchar idata card_snr[4]; //配置字 uchar data bankdata[28] = {1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7}; //存儲卡上用戶數據(1-7)7*4=28 uchar data cominceptbuff[6] = {1,2,3,4,5,6};//串口接收數組ram uchar command; //第一個命令 uchar command1;// //uint temp; uchar j,i; uchar myaddr = 8; //uchar ywqz_count,time_count; //ywqz jishu: uchar bdata DATA; sbit BIT0 = DATA^0; sbit BIT1 = DATA^1; sbit BIT2 = DATA^2; sbit BIT3 = DATA^3; sbit BIT4 = DATA^4; sbit BIT5 = DATA^5; sbit BIT6 = DATA^6; sbit BIT7 = DATA^7; uchar bdata DATA1; sbit BIT10 = DATA1^0; sbit BIT11 = DATA1^1; sbit BIT12 = DATA1^2; sbit BIT13 = DATA1^3; sbit BIT14 = DATA1^4; sbit BIT15 = DATA1^5; sbit BIT16 = DATA1^6; sbit BIT17 = DATA1^7; bit i_CurrentLevel;//i_CurrentLevel BIT 00H(Saves current level of OutPut pin of U2270B) bit timer1_end; bit read_ok = 0; //緩存定時值,因用同一個定時器 union HLint { uint W; struct { uchar H;uchar L; } B; };//union HLint idata a union HLint data a; //緩存定時值,因用同一個定時器 union HLint0 { uint W; struct { uchar H; uchar L; } B; };//union HLint idata a union HLint0 data b; /**********************函數原型*****************/ //讀寫操作 void f_readcard(void);//全部讀出1~7 AOR喚醒 void f_writecard(uchar x);//根據命令寫不同的內容和操作 void f_clearpassword(void);//清除密碼 void f_changepassword(void);//修改密碼 //功能子函數 void write_password(uchar data *data p);//寫初始密碼或數據 void write_block(uchar x,uchar data *data p);//不能用通用指針 void write_bit(bit x);//寫位 /*子函數區*****************************************************/ void delay_2(uint x) //延時,時間x*10us@12mhz,最小20us@12mhz { x--; x--; while(x) { _nop_(); _nop_(); x--; } _nop_();//WDT_CONTR=0X3C;不能頻繁的復位 _nop_(); } ///////////////////////////////////////////////////////////////////// void initial(void) { SCON = 0x50; //串口方式1,允許接收 //SCON =0x50; //01010000B:10位異步收發,波特率可變,SM2=0不用接收到有效停止位才RI=1, //REN=1允許接收 TMOD = 0x21; //定時器1 定時方式2(8位),定時器0 定時方式1(16位) TCON = 0x40; //設定時器1 允許開始計時(IT1=1) TH1 = 0xfD; //FB 18.432MHz 9600 波特率 TL1 = 0xfD; //fd 11.0592 9600 IE = 0X90; //EA=ES=1 TR1 = 1; //啟動定時器 WDT_CONTR = 0x3c;//使能看門狗 p_U2270B_Standby = 0;//單電源 PCON = 0x00; IP = 0x10;//uart you xian XXXPS PT1 PX1 PT0 PX0 led_light1 = 1; led_light = 0; p_U2270B_OutPut = 1; } /************************************************/ void f_readcard()//讀卡 { EA = 0;//全關,防止影響跳變的定時器計時 WDT_CONTR = 0X3C;//喂狗 p_U2270B_CFE = 1;// delay_2(232); //>2.5ms /* // aor 用喚醒功能來防碰撞 p_U2270B_CFE = 0; delay_2(18);//start gap>150us write_bit(1);//10=操作碼讀0頁 write_bit(0); write_password(&bankdata[24]);//密碼block7 p_U2270B_CFE =1 ;// delay_2(516);//編程及確認時間5.6ms */ WDT_CONTR = 0X3C;//喂狗 led_light = 0; b.W = 0; while(!(read_ok == 1)) { //while(p_U2270B_OutPut);//等一個穩定的低電平?超時判斷? while(!p_U2270B_OutPut);//等待上升沿的到來同步信號檢測1 TR0 = 1; //deng xia jiang while(p_U2270B_OutPut);//等待下降沿 TR0 = 0; a.B.H = TH0; a.B.L = TL0; TH0 = TL0 = 0; TR0 = 1;//定時器晚啟動10個周期 //同步頭 if((324 < a.W) && (a.W < 353)) ;//檢測同步信號1 else { TR0 = 0; TH0 = TL0 = 0; goto read_error; } //等待上升沿 while(!p_U2270B_OutPut); TR0 = 0; a.B.H = TH0; a.B.L = TL0; TH0 = TL0 = 0; TR0 = 1;//b.N1<<=8; if(a.B.L < 195);//0.5p else { TR0 = 0; TH0 = TL0 = 0; goto read_error; } //讀0~7塊的數據 for(j = 0;j < 28;j++) { //uchar i; for(i = 0;i < 16;i++)//8個位 { //等待下降沿的到來 while(p_U2270B_OutPut); TR0 = 0; a.B.H = TH0; a.B.L = TL0; TH0 = TL0 = 0; TR0 = 1; if(t2_max < a.W/*)&&(a.W < t2_max)*/)//1P { b.W >>= 2;//先左移再賦值 b.B.L += 0xc0; i++; } else if(t1_min < a.B.L/*)&&(a.B.L < t1_max)*/)//0.5p { b.W >>= 1; b.B.L += 0x80; } else { TR0 = 0; TH0 = TL0 = 0; goto read_error; } i++; while(!p_U2270B_OutPut);//上升 TR0 = 0; a.B.H = TH0; a.B.L = TL0; TH0 = TL0 = 0; TR0 = 1; if(t2_min < a.W/*)&&(a.W < t2_max)*/)//1P { b.W >>= 2; i++; } else if(t1_min < a.B.L/*a.W)&&(a.B.L < t1_max)*/)//0.5P //else if(!(a.W==0)) { b.W >>= 1; //temp+=0x00; //led_light1=0;led_light=1;delay_2(40000); } else { TR0 = 0; TH0 = TL0 = 0; goto read_error; } i++; } //取出奇位 DATA = b.B.L; BIT13 = BIT7; BIT12 = BIT5; BIT11 = BIT3; BIT10 = BIT1; DATA = b.B.H; BIT17 = BIT7; BIT16 = BIT5; BIT15 = BIT3; BIT14 = BIT1; bankdata[j] = DATA1; } read_ok = 1;//讀卡完成了 read_error: _nop_(); } } /***************************************************/ void f_writecard(uchar x)//寫卡 { p_U2270B_CFE = 1; delay_2(232); //>2.5ms //psw=0 standard write if (x == write_command0)//寫密碼:初始化密碼 { uchar i; uchar data *data p; p = cominceptbuff; p_U2270B_CFE = 0; delay_2(31);//start gap>330us write_bit(1);//寫操作碼1:10 write_bit(0);//寫操作碼0 write_bit(0);//寫鎖定位0 for(i = 0;i < 35;i++) { write_bit(1);//寫數據位1 } p_U2270B_CFE = 1; led_light1 = 0; led_light = 1; delay_2(40000);//測試使用 //write_block(cominceptbuff[4],p); p_U2270B_CFE = 1; bankdata[20] = cominceptbuff[0];//密碼存入 bankdata[21] = cominceptbuff[1]; bankdata[22] = cominceptbuff[2]; bankdata[23] = cominceptbuff[3]; } else if (x == write_command1)//配置卡參數:初始化 { uchar data *data p; p = cominceptbuff; write_bit(1);//寫操作碼1:10 write_bit(0);//寫操作碼0 write_bit(0);//寫鎖定位0 write_block(cominceptbuff[4],p); p_U2270B_CFE= 1; } //psw=1 pssword mode else if(x == write_command2) //密碼寫數據 { uchar data*data p; p = &bankdata[24]; write_bit(1);//寫操作碼1:10 write_bit(0);//寫操作碼0 write_password(p);//發口令 write_bit(0);//寫鎖定位0 p = cominceptbuff; write_block(cominceptbuff[4],p);//寫數據 } else if(x == write_command3)//aor //喚醒 { //cominceptbuff[1]操作碼10 X xxxxxB uchar data *data p; p = cominceptbuff; write_bit(1);//10 write_bit(0); write_password(p);//密碼 p_U2270B_CFE = 1;//此時數據不停的循環傳出 } else //停止操作碼 { write_bit(1);//11 write_bit(1); p_U2270B_CFE = 1; } p_U2270B_CFE = 1; delay_2(560);//5.6ms } /************************************/ void f_clearpassword()//清除密碼 { uchar data *data p; uchar i,x; p = &bankdata[24];//原密碼 p_U2270B_CFE = 0; delay_2(18);//start gap>150us //操作碼10:10xxxxxxB write_bit(1); write_bit(0); for(x = 0;x < 4;x++)//發原密碼 { DATA = *(p++); for(i = 0;i < 8;i++) { write_bit(BIT0); DATA >>= 1; } } write_bit(0);//鎖定位0:0 p = &cominceptbuff[0]; write_block(0x00,p);//寫新配置參數:pwd=0 //密碼無效:即清除密碼 DATA = 0x00;//停止操作碼00000000B for(i = 0;i < 2;i++) { write_bit(BIT7); DATA <<= 1; } p_U2270B_CFE = 1; delay_2(560);//5.6ms } /*********************************/ void f_changepassword()//修改密碼 { uchar data *data p; uchar i,x,addr; addr = 0x07;//block7 p = &Nkey_a[0];//原密碼 DATA = 0x80;//操作碼10:10xxxxxxB for(i = 0;i < 2;i++) { write_bit(BIT7); DATA <<= 1; } for(x = 0;x < 4;x++)//發原密碼 { DATA = *(p++); for(i = 0;i < 8;i++) { write_bit(BIT7); DATA >>= 1; } } write_bit(0);//鎖定位0:0 p = &cominceptbuff[0]; write_block(0x07,p);//寫新密碼 p_U2270B_CFE = 1; bankdata[24] = cominceptbuff[0];//密碼存入 bankdata[25] = cominceptbuff[1]; bankdata[26] = cominceptbuff[2]; bankdata[27] = cominceptbuff[3]; DATA = 0x00;//停止操作碼00000000B for(i = 0;i < 2;i++) { write_bit(BIT7); DATA <<= 1; } p_U2270B_CFE = 1; delay_2(560);//5.6ms } /***************************子函數***********************************/ void write_bit(bit x)//寫一位 { if(x) { p_U2270B_CFE = 1; delay_2(32);//448*11.0592/120=42延時448us p_U2270B_CFE = 0; delay_2(28);//280*11.0592/120=26寫1 } else { p_U2270B_CFE = 1; delay_2(92);//192*11.0592/120=18 p_U2270B_CFE = 0; delay_2(28);//280*11.0592/120=26寫0 } } /*******************寫一個block*******************/ void write_block(uchar addr,uchar data *data p) { uchar i,j; for(i = 0;i < 4;i++)//block0數據 { DATA = *(p++); for(j = 0;j < 8;j++) { write_bit(BIT0); DATA >>= 1; } } DATA = addr <<= 5;//0地址 for(i = 0;i < 3;i++) { write_bit(BIT7); DATA <<= 1; } } /*************************************************/ void write_password(uchar data *data p) { uchar i,j; for(i = 0;i < 4;i++)// { DATA = *(p++); for(j = 0;j < 8;j++) { write_bit(BIT0); DATA >>= 1; } } } /*************************************************/ void main() { initial(); TI = RI = 0; ES = 1; EA = 1; delay_2(28); //f_readcard(); while(1) { f_readcard(); //讀卡 f_writecard(command1); //寫卡 f_clearpassword(); //清除密碼 f_changepassword(); //修改密碼 } }
標簽： 12345
上傳時間： 2017-10-20
上傳用戶：my_lcs
題目：古典問題：有一對兔子，從出生后第3個月起每個月都生一對兔子，小兔子長到第三個月后每個月又生一對兔子，假如兔子都不死，問每個月的兔子總數為多少？ //這是一個菲波拉契數列問題 public class lianxi01 { public static void main(String[] args) { System.out.println("第1個月的兔子對數: 1"); System.out.println("第2個月的兔子對數: 1"); int f1 = 1, f2 = 1, f, M=24; for(int i=3; i<=M; i++) { f = f2; f2 = f1 + f2; f1 = f; System.out.println("第" + i +"個月的兔子對數: "+f2); } } } 【程序2】 題目：判斷101-200之間有多少個素數，并輸出所有素數。 程序分析：判斷素數的方法：用一個數分別去除2到sqrt(這個數)，如果能被整除， 則表明此數不是素數，反之是素數。 public class lianxi02 { public static void main(String[] args) { int count = 0; for(int i=101; i<200; i+=2) { boolean b = false; for(int j=2; j<=Math.sqrt(i); j++) { if(i % j == 0) { b = false; break; } else { b = true; } } if(b == true) {count ++;System.out.println(i );} } System.out.println( "素數個數是: " + count); } } 【程序3】 題目：打印出所有的 "水仙花數 "，所謂 "水仙花數 "是指一個三位數，其各位數字立方和等于該數本身。例如：153是一個 "水仙花數 "，因為153=1的三次方＋5的三次方＋3的三次方。 public class lianxi03 { public static void main(String[] args) { int b1, b2, b3;
上傳時間： 2017-12-24
上傳用戶：Ariza
