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Description: 第1章 文件结构... 11 1.1 版权和版本的声明... 11 1.2 头文件的结构... 12 1.3 定义文件的结构... 13 1.4 头文件的作用... 13 1.5 目录结构... 14 第2章 程序的版式... 15 2.1 空行... 15 2.2 代码行... 16 2.3 代码行内的空格... 17 2.4 对齐... 18 2.5 长行拆分... 19 2.6 修饰符的位置... 19 2.7 注释... 20 2.8 类的版式... 21 第3章 命名规则... 22 3.1 共性规则... 22 3.2 简单的Windows应用程序命名规则... 23 3.3 简单的Unix应用程序命名规则... 25 第4章 表达式和基本语句... 26 4.1 运算符的优先级... 26 4.2 复合表达式... 27 4.3 if 语句... 27 4.4 循环语句的效率... 29 4.5 for 语句的循环控制变量... 30 4.6 switch语句... 30 4.7 goto语句... 31 第5章 常量... 33 5.1 为什么需要常量... 33 5.2 const 与 #define的比较... 33 5.3 常量定义规则... 33 5.4 类中的常量... 34 第6章 函数设计... 36 6.1 参数的规则... 36 6.2 返回值的规则... 37 6.3 函数内部实现的规则... 39 6.4 其它建议... 40 6.5 使用断言... 41 6.6 引用与指针的比较... 42 第7章 内存管理... 44 7.1内存分配方式... 44 7.2常见的内存错误及其对策... 44 7.3指针与数组的对比... 45 7.4指针参数是如何传递内存的？... 47 7.5 free和delete把指针怎么啦？... 50 7.6 动态内存会被自动释放吗？... 50 7.7 杜绝“野指针”... 51 7.8 有了malloc/free为什么还要new/delete ？... 52 7.9 内存耗尽怎么办？... 53 7.10 malloc/free 的使用要点... 54 7.11 new/delete 的使用要点... 55 7.12 一些心得体会... 56 第8章 C++函数的高级特性... 57 8.1 函数重载的概念... 57 8.2 成员函数的重载、覆盖与隐藏... 60 8.3 参数的缺省值... 63 8.4 运算符重载... 64 8.5 函数内联... 65 8.6 一些心得体会... 68 第9章 类的构造函数、析构函数与赋值函数... 69 9.1 构造函数与析构函数的起源... 69 9.2 构造函数的初始化表... 70 9.3 构造和析构的次序... 72 9.4 示例：类String的构造函数与析构函数... 72 9.5 不要轻视拷贝构造函数与赋值函数... 73 9.6 示例：类String的拷贝构造函数与赋值函数... 73 9.7 偷懒的办法处理拷贝构造函数与赋值函数... 75 9.8 如何在派生类中实现类的基本函数... 75 9.9 一些心得体会... 77 第10章 类的继承与组合... 78 10.1 继承... 78 10.2 组合... 80 第11章 其它编程经验... 82 11.1 使用const提高函数的健壮性... 82 11.2 提高程序的效率... 84 11.3 一些有益的建议
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Author: 87503655@qq.com |
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Description: ADI公司的关于光通讯模块的监控程序,在KEIL FOR ARM 的编译环境编译.程序大小约12K,芯片是ADu7020.-ADI on the optical communication module procedures of control in KEIL FOR ARM compiler compiler environment. About 43-45 procedures size, the chip is ADu7020.
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Size: 291840 |
Author: 张涛 |
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Description: 本游戏是采用斜45度角的角色扮演游戏。游戏以一个想象的世界为背景，以一个梦想成为英雄的小虾米锦霜为主角，产开一幅庞大的故事画面。在锦霜的冒险历程中，认识了两位女主角：雪一般的忆雪和火一般的红怜，从而产开了一段刻骨铭心的爱情故事。这只是我们游戏的一个部分，或者说是一条主线。随着主角们冒险历程的不断深入，随着故事的不断发展，红狼、素心儿、恒卿、影枫、寒月等一批有血有肉的角色不断加入，各自书写着各自独特的人生，各自独特的悲剧。游戏中，黑暗组织的神秘面纱慢慢被揭开，他们酝酿的大阴谋是什么？在黑暗组织的背后，似乎还有一股更为强大的力量，还有一个更大的阴谋，那是什么？隐藏在忆雪身上的秘密是什么？随着我们故事的发展，一切都会揭开。
-the game is tilted 45 degrees angle of role-playing games. Imagine a game in the world as a backdrop to a dream become a hero of the outgoing Kam cream for the leading role, a huge open-story screen. At Kam cream adventure course, understand the two Actress : Ice general Recalling snow and fiery red Pity, thereby opening the middle section of unforgettable love story. It's just a part of the game, or is a main line. With the main players in the adventures continued to deepen, as the story continues to develop, the referee, Concolor abuse, constant Betty, videos maples firm flesh of a number of roles constantly, each with its own unique writing life, each unique tragedy. Games, the dark veil of secrecy has been lifted slowly, the big brewing conspiracy? In the darkness behind, it seems
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Size: 707584 |
Author: 刘坤 |
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Description: 软件版本：
V0.4
发布日期：
23:43 2005-10-9
运行环境：
MS-DOS|Windows 9x/ME|Windows XP
软件简介：
这是用NEO V1.1.45写的打砖块游戏。主架构从网友的作品中移植而来，NEO被用来增强性能和美化界面。由于使用的是DOS接口，所以可能在部分AGP 8X的显卡下会出现黑屏。
规则说明：
和普通的打砖块游戏差不多，鼠标操作。一共有5关，速度越来越快，要打爆是还是很难的。
联系我们：
ckerhome@yahoo.com.cn-Software Version : V0.4 release date : 23:43 2005-10-9 runtime environment : MS-DOS | Windows Editions | Windows XP Introduction : This is used to write the NEO V1.1.45 game arcade. Netizens from the main framework of the works are transplanted, NEO be used to enhance performance and landscaping interface. As the use of DOS interface, it may in some AGP 8X graphics will emerge under a black screen. Rules : general and the almost arcade games, mouse operation. A total of five customs, increasingly faster, or to be bombarded very difficult. Contact : ckerhome@yahoo.com.cn
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Size: 63488 |
Author: 董凯 |
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Description: 解决时钟问题，acm竞赛题
A weird clock marked from 0 to 59 has only a minute hand. It won t move until a special coin is thrown into its box. There are different kinds of coins as your options. However once you make your choice, you cannot use any other kind. There are infinite number of coins of each kind, each marked with a number d ( 0 <= 1000 ), meaning that this coin will make the minute hand move d times clockwise the current time. For example, if the current time is 45, and d = 2. Then the minute hand will move clockwise 90 minutes and will be pointing to 15.
Now you are given the initial time s ( 0 <= s <= 59 ) and the coin s type d. Write a program to find the minimum number of d-coins needed to turn the minute hand back to 0.-clock to solve the problem, acm contest A weird clock marked from 0 to 59 has only a minute hand. It won t move until a special coin is thrown into its box. There are different kinds of coins as your options. However once you make your choice, you can not use any other kind. There are infinite number of coins of each kind, each marked with a number d (0 lt; = 1000), meaning that this coin will make the minute hand move clockwise d times the current time. For example, if the current time is 45, and d = 2. Then the minute hand will move clockwise 90 minutes and will be pointing to 15. Now you are given the initial time s (0 lt; s = lt; = 59) and the coin s type d. Write a program to find the minimum number of d-coins needed to turn the minute hand back to 0.
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Size: 1024 |
Author: 王冠 |
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Description: /*红外解码的方法 NEC格式 upd6121
1、9Ms的高电平启动头，然后是4.5Ms的低电平，如果2.25Ms时就有高电平，是持续信号，不处理
2、然后以一个高电平和一个低电平为1Bit，高电平时间是0.5625Ms=562us，
高低电平时间比为1:1时是Bit1，比为1:3时是Bit0
3、Timer1中断执行100us采样周期，9Ms＝90，4.5Ms=45，高电平＝5，低电平最多15
4、共读入4Byte共24bit，第1、2Byte是CustomCode码和其反码，第3、4Byte是DataCode和其反码
5、CustomCode正确和DataCode效验正确的话，执行
6、红外接收器输出是反相的-/* IR decoding methods NEC format upd61211, 9Ms start of the high head, and then the low 4.5ms, if 2.25Ms when there is high, is a continuous signal, does not handle 2, and then to a high- level and a low level for 1Bit, high time is 0.5625Ms = 562us, high-low ratio of 1:1 when the time is Bit1, ratio of 1:3 is Bit03, Timer1 interrupt the implementation of 100us sampling period, 9Ms = 90,4.5 Ms = 45, high = 5, low-level maximum of 154, were read into 4Byte were 24bit, No. 1,2 Byte is CustomCode code and its anti-code, the first 3,4 Byte is DataCode and its anti-code 5, CustomCode correct and well-tested DataCode correct, the implementation of 6, infrared receiver output is RP s
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Size: 692224 |
Author: 王宇 |
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Description: 全班有N（2<=N<=45）个人排成一排，但因为高矮不齐，需要进行调整。调整的方法是，不调换左右次序，只让若干人后退一步变为第2排，使第一排留下的人从左到右的身高按降序排列，即右边的人不比左边的人高。如果第2排的人还不按降序排列，则照此办理，即再让第2排的若干人后退一步变为第3排，这样继续下去，直到所有排的人都按身高从高到低排列。
调整中，你需要找出一种使第一排留下的人数尽可能多的调整方法，第二排若需要继续调整，则也应使第二排留下的人数尽可能多，余类推。
-school there N (2
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Size: 2048 |
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Description: This a GBA(Game Boy Advance) animation sample code. It continue and reverse display 45 BMPs on GBA screen let it looks like an animation.-This a GBA (Game Boy Advance)'s animation ample code. It continue and reverse display 45 B BMPs on GBA screen let it looks like an animation.
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Size: 1253376 |
Author: Aaron99 |
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Description: 当然他好通惠家园进45一任何牛肉汤教育局人格温热图5一45 王汝刚奖金额汤旺河安徽认为他易货交易精华-Of course, good通惠家园him into any one of 45 beef soup warm personality Fig.5 Education 1 45王汝刚prize money Tangwang River in Anhui consider him the essence of barter transactions
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Size: 82944 |
Author: lhh321 |
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Description: 基于 LabWindowsCVI 的虚拟仪器设计与应用45-LabWindowsCVI Based on Virtual Instrument Design and Application of 45
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Size: 5120 |
Author: yangliang |
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Description: 红外解码的方法 NEC格式 upd6121
1、9Ms的高电平启动头，然后是4.5Ms的低电平，如果2.25Ms时就有高电平，是持续信号，不处理
2、然后以一个高电平和一个低电平为1Bit，高电平时间是0.5625Ms=562us，
高低电平时间比为1:1时是Bit1，比为1:3时是Bit0
3、Timer1中断执行100us采样周期，9Ms＝90，4.5Ms=45，高电平＝5，低电平最多15
4、共读入4Byte共24bit，第1、2Byte是CustomCode码和其反码，第3、4Byte是DataCode和其反码
5、CustomCode正确和DataCode效验正确的话，执行
6、红外接收器输出是反相的-Infrared NEC format decoding method upd61211, 9Ms start of the high head, and then the low 4.5ms, if 2.25Ms when there is high, is a continuous signal, do not deal with 2, and then to a high and a low level for 1Bit, high time is 0.5625Ms = 562us, high-low ratio of 1:1 when the time is Bit1, ratio of 1:3 is Bit03, Timer1 interrupt the implementation of 100us sampling period, 9Ms = 90, 4.5Ms = 45, high = 5, low-level maximum of 154, were read into 4Byte were 24bit, No. 1,2 Byte is CustomCode code and its anti-code, the first is the 3,4 Byte Code DataCode and its anti-5, CustomCode correct and well-tested DataCode correct, the implementation of 6, infrared receiver output is RP s
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Size: 821248 |
Author: wang |
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Description: int gdriver=DETECT, gmode, errorcode
struct arccoordstype arcinfo
int midx, midy
int stangle=45,endangle=270
char sstr[80],estr[80] -int gdriver = DETECT, gmode, errorcode struct arccoordstype arcinfo int midx, midy int stangle = 45, endangle = 270 char sstr [80], estr [80]
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Size: 1024 |
Author: ly |
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Description: 三点A，B，C (坐标A.x, A.y ......)
A、B是矩形的两端点
且C在与AB平行的 矩形另一边上 或其延长线上
求矩形另外两端点M、N
再求与MN夹角为45度，与M距离为10的点P
以及与MN夹角45度，与N距离为10的点Q
注：P、Q都在矩形内部
-Three points A, B, C (coordinates Ax, Ay ......) A, B is rectangular and the two end points of C with AB parallel to the edge of another rectangle or its extension line of the other two end points for the rectangular M, N re-seek with the MN angle of 45 degrees, and M a distance of 10 points, P and MN angle 45 degrees, and N a distance of 10-point Q Note: P, Q are inside the rectangular
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Size: 1024 |
Author: 李智信 |
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Description: C语言精彩百例第45－70例
第二篇 深入提高篇
实例45 结构体变量
实例46 结构体数组
实例47 结构体指针变量
实例48 结构体指针数组
实例49 共用体变量
实例50 枚举类型
实例51 读写字符
实例52 读写字符串
实例53 格式化输出函数
实例54 格式化输入函数
实例55 打开和关闭文件
实例56 fputc()和fgetc()
实例57 函数rewind()
实例58 fread()和fwrite()
实例59 fprintf()和fscanf()
实例60 随机存取
实例61 错误处理
实例62 综合实例
实例63 动态分配函数
实例64 常用时间函数
实例65 转换函数
实例66 查找函数
实例67 跳转函数
实例68 排序函数
实例69 伪随机数生成
实例70 可变数目变元-err
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Size: 22528 |
Author: 杨志亮 |
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Description: 一个实用的单片机液晶多级菜单例子，用keilc编译器运行-Single-chip LCD, a practical example of multi-level menu, use the compiler to run keilc
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Size: 12288 |
Author: Jerry |
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Description: C51实用程序集（45个）直接下载
C51实用程序集（45个）直接下载-C51 utility sets (45) Direct download C51 utility sets (45) Direct Download
Platform: |
Size: 531456 |
Author: ltj |
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Description: 1、用RS－232C的零MODEM的连接法连好两机的连线；
2、用DEBUG的O命令将一数发送到串行口[2F8H/3F8H]，然后在另一台机上用 I命令读入串行口的内容。如果发送的的内容与接受的内容一直，说明两机已经连好线，否则必须检查连线，直到两机可通讯为止；
3、编一程序，将一台机的键盘输入的键值通过RS-232C口送到另一台机，并在CRT上显示键值；
4、编一程序，通过RS-232C口进行一台机到另一台机的文件复制。-1, using RS-232C connection zero MODEM good law even connect the two machines 2, using the O command DEBUG number will be sent to one serial port [2F8H/3F8H], then on another machine I use the command Read the contents into the serial port. If you send the content and the content has been accepted that two planes have been even better line, he must check the connection, until the two planes can be so communication 3, made one procedure will be a machine keyboard keys through the RS-232C port to another machine, and CRT display keys 4, Addendum 1, through RS-232C port for a machine to another machine to copy documents.
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Size: 56320 |
Author: 刘金来 |
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Description: 2.45GHZ有源标签的源代码，主要是MSP430控制射频模块的收发-2.45GHZ active tag' s source code, the main control are MSP430 RF transceiver modules
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Size: 43008 |
Author: 付琪 |
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Description: 单片机c语言小程序，45个C51实用程序集 我也是下载的，希望对学习者有用，版权归原作者，其他用途请向原作者申请。-Small single-chip c language program, 45 C51 utility set I downloaded, and I hope useful for learners, belongs to original author, for other purposes please apply to the original author.
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Size: 531456 |
Author: 康生 |
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Description: 计算序列的卷积和相关函数 1. 计算序列h(n)和x(n)的卷积，h(n)*x(n)。 //
// //
// 2. 计算序列h(n)和h(n-45)的相关函数。-DSP
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Size: 1024 |
Author: grdffa |
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