-
5
说明: 本书作者根据自己学习C++的亲身体会及多年教学经验,用简单的例子和简练的叙述讲解C++编程,别具特色。
全书共分十八章(此为第10到12章),内容涉及对象的演化、数据抽象、隐藏实现、初始化与清除、函数重载与缺省参数、输入输出流介绍、常量、内联函数、命名控制、引用和拷贝构造函数、运算符重载、动态对象创建、继承和组合、多态和虚函数、模板和包容器类、多重继承、异常处理和运行时类型识别。
本书作为正式教材和自学用书均非常优秀,作为程序设计者的参考用书亦极为合适。
(Of this book to learn C++ based on their personal experience and years of teaching experience, using simple examples and concise narrative explaining C++ programming, unique.
The book is divided into eighteen chapters, covering the evolution of objects, data abstraction, hiding implementation, initialization and removal, function overloading and default parameters, input and output streams introduced constants, inline functions, named control, reference and copy constructors, operator overloading, dynamic object creation, inheritance and composition, polymorphism and virtual functions, templates, and the container classes, multiple inheritance, exception handling and runtime type identification.
Book as an official textbook, and a self-learning books are very good, as the programmer s reference book is also very appropriate.)
- 2010-12-14 19:20:09下载
- 积分:1
-
example
有400个C#和.net方面的源码和源码说明(400 C# and. Net source and source Description)
- 2013-02-07 12:00:33下载
- 积分:1
-
419089Demo
说明: 摄像头 低压低噪马达驱动--MS41908(Camera Low Voltage and Low Noise Motor Drive--MS41908)
- 2020-06-16 07:20:01下载
- 积分:1
-
C Sharp Windows Forms 程序设计
一本比较好的Winform 的书,使用C Sharp语言(One good book for winform with C#.)
- 2020-06-23 08:20:02下载
- 积分:1
-
lesson01.tar
qt教学例子1,qiliang 所写,适于初学者使用,版本3.0
(qt an example of teaching, qiliang wrote, suitable for beginners, Version 3.0)
- 2007-02-28 17:40:21下载
- 积分:1
-
C++ACC编码程序
一个C++ACC编码程序,里面包含音频ACC编码库的程序,全部都是用C++和C实现的,有需要的可以下载来看下....
- 2022-05-29 07:50:34下载
- 积分:1
-
111OS
说明: UCOS2最小内核分析手册,包含OS操作函数说明等详细内容(UCOS2 minimum core of manual operation function that contains the OS and other details)
- 2011-02-20 18:34:29下载
- 积分:1
-
Flight_System
用MFC做的航空售票系统,是数据库课程设计,全部是自己做的,还不错哦。(MFC to do with air ticketing system is a database of curriculum design, to do all of their own, oh well.)
- 2009-04-02 21:18:58下载
- 积分:1
-
medo
设X[ 0 : n - 1]和Y[ 0 : n – 1 ]为两个数组,每个数组中含有n个已排好序的数。找出X和Y的2n个数的中位数。 编程任务 利用分治策略试设计一个O (log n)时间的算法求出这2n个数的中位数。 数据输入 由文件input.txt提供输入数据。文件的第1行中有1个正整数n(n<=200),表示每个数组有n个数。接下来的两行分别是X,Y数组的元素。结果输出 程序运行结束时,将计算出的中位数输出到文件output.txt中(Let X [0: n- 1] and Y [0: n- 1] for the two arrays, each array containing the n number has been sorted. 2n X and Y to identify the number of digits. programming tasks using the divide and conquer strategy try to design an O (log n) time algorithm to calculate this median number 2n. Data input by the input data provided input.txt file. The first line in the file has a positive integer n (n < = 200), that there are n numbers of each array. The next two lines are the X, Y array elements. The end result is output program runs, the calculated median output to file output.txt)
- 2021-03-22 16:29:16下载
- 积分:1
-
ER
复杂网络研究中的,ER模型图的生成程序,使用C++写成。(ER model diagram generation program in the study of complex networks, C++ languages.)
- 2013-05-15 19:55:37下载
- 积分:1