UVCView.x86

VC++ 源代码 文档 共10节 第一节(VC source code files a total of 10 Section I)

说明：  labview调用dll实现QR二维码生成，包含dll说明等，可作为子vi直接使用-LabVIEW(calls the DLL implementation of QR code generation, including DLL, can be directly used as a sub VI)

利用C++实现的一个即时通讯系统的客户端。传输层采用的是TCP协议。这是本科时候的毕业设计，最后评定为优，希望对一些朋友能有所帮助。(The use of C++ implementation of an instant messaging system client. Transport layer protocol used is TCP. This is the time of undergraduate graduation project, and finally classified as gifted, in the hope that some of my friends can be helpful.)

用VC编写的媒体播放器，虽然功能不是很强大，可以供大家参考(VC prepared by the Media Player, although the function is not very strong, for your reference)

usbview 在VS2010的编译通过的实例(usbview build in VS2010)

Windows2000/xp下plx9054的pci驱动（源码、工程文件），支持DMA操作。 使用ddk开发，工程可以直接编译。(plx9054 Driver under Windows2000/XP. Developed by DDK.)

屏幕录制程序 联合开发网 - pudn.com

说明：  最简单的USB通讯上位机程序，实现发送和接收。(The simplest PC USB communications process, the achievement of sending and receiving.)

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)

播放器的搭建和操作，精美播放器界面设置...............(Build and operate the player, fine player interface settings )