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uCOS
uCOS-II_LPC2000移植源代码,可移植到基于LPC2000系列芯片的嵌入式系统-uCOS-II_LPC2000 transplant source code, can be transplanted into series based on the LPC2000 chip, embedded systems
- 2023-09-03 04:50:03下载
- 积分:1
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pthread_t created with input from the standard input of the number N of N thread...
用pthread_t创建由标准输入端输入个数N的N个线程,分别计算各自的结果,再返回给主进程.-pthread_t created with input from the standard input of the number N of N threads, calculate their results and then back to the main process.
- 2022-07-03 04:29:20下载
- 积分:1
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使用串行通信网络
linux下利用串口进行网络通讯-using serial communications network
- 2022-07-15 18:58:18下载
- 积分:1
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Nemesis 2D Soccer Simulation Team
RoboCup 2008 2D仿真组 Nemesis队 源代码-Nemesis 2D Soccer Simulation Team
- 2022-02-01 09:35:00下载
- 积分:1
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uclinux下串口通信例子,实现基本的收发功能
uclinux下串口通信例子,实现基本的收发功能-Examples of serial communication under uClinux, the realization of basic transceiver functions
- 2022-01-21 19:56:35下载
- 积分:1
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设备驱动程序开发,一个清晰版的第三版,很理想…
linux设备驱动开发,清晰版本第三版,非常适合LINUX开发驱动使用-linux device driver development, a clear version of the third edition, is ideal for the development of drivers to use LINUX
- 2022-04-30 17:19:22下载
- 积分:1
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Well
著名图论问题之一。邮递员从邮局出发送信,要求对辖区内每条街,都至少通过一次,再回邮局。在此条件下,怎样选择一条最短路线?此问题由中国数学家管梅谷于1960年首先研究并给出算法,故名。-Well-known graph theory problems. Postman deliver a letter from the post office proceed to require every street within the right area, at least through the time, return to post office. Under these conditions, how to select a shortest route? This issue by the Chinese mathematician Mei Gu tube study in 1960, and gives the first algorithm, hence the name.
- 2022-06-18 09:20:13下载
- 积分:1
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网络编程,uc/os下用的IP/TCp协议,嵌入式开发必备
网络编程,uc/os下用的IP/TCp协议,嵌入式开发必备-network programming, uc/os using IP/TCp agreement, embedded development required
- 2022-02-22 05:52:53下载
- 积分:1
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n皇后
n皇后-n Queen
- 2022-04-11 08:18:50下载
- 积分:1
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linux 2.6内核版本的设备驱动详解书籍,对linux设备驱动的编写很有用!...
linux 2.6内核版本的设备驱动详解书籍,对linux设备驱动的编写很有用!-linux 2.6 kernel version of the device driver Xiangjie books, linux device driver for the preparation of very useful!
- 2022-01-26 07:54:46下载
- 积分:1