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sipp 一个不错的SIP代理或软交换测试工具
sipp 一个不错的SIP代理或软交换测试工具-SIPP a good SIP proxy or softswitch testing tools
- 2022-02-14 15:56:36下载
- 积分:1
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Qt4编程指南,“C++ GUI QT4编程与“中国版…
Qt4程序设计指南,《C++ GUI Programming with Qt4 》的中文版,通过一系列简单的实例,让初学者尽快了解Qt编程,是一本不可多得的好书-Qt4 Programming Guide, " C++ GUI Programming with Qt4" the Chinese version, through a series of simple example, as soon as possible so that beginners learn Qt programming, is a rare book
- 2022-05-18 09:26:12下载
- 积分:1
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在Linux环境下播放器mplayer的源
在Linux环境下的Mplayer播放器源码-In the Linux environment player Mplayer source
- 2022-02-20 13:39:54下载
- 积分:1
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一组GUI教学程序
一组GUI教学程序-a group GUI teaching procedures
- 2022-03-24 19:12:39下载
- 积分:1
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under linux the most important one of VIM editor of the 6.03 version of the Chin...
linux下最重要的编辑器之一VIM的6.03版中文手册-under linux the most important one of VIM editor of the 6.03 version of the Chinese manual
- 2022-08-06 12:30:56下载
- 积分:1
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Linux系统环境下,设备驱动程序开发入门,作为参考。
Linux系统环境下,设备驱动程序开发入门,作为参考。-Linux system environment, the device driver entry, as a reference.
- 2022-02-21 01:58:35下载
- 积分:1
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内核定时器的使用+linux+timer
内核定时器的使用+linux+timer-The use of the kernel timers+ linux+ timer
- 2022-07-01 06:46:22下载
- 积分:1
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该模拟的UNIX文件系统应该有以下功能:
多级目录的结构,第一目录为用户名,其余的为它的子目录;
多用户的功能,有8个用户使用同一块数据区;
具有...
该模拟的UNIX文件系统应该有以下功能:
多级目录的结构,第一目录为用户名,其余的为它的子目录;
多用户的功能,有8个用户使用同一块数据区;
具有用户登录和注册的功能,如果用户没有用户名和密码就应该注册,如果有了用户名和密码就应该登陆;
-simulation of the UNIX file system should have the following features : multi-level directory structure, the first directory to the user name, the rest are in a subdirectory; Multi-user functions, eight users with a data area; With user login and registration functions, if users do not have a user name and secret the codes would be registered, if we have a user name and password should landing;
- 2022-07-21 08:38:49下载
- 积分:1
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linux programming simple example, covering file I/O, multi
linux编程的简单例子,涵盖文件I/O,多线程,多进程,socket等入门经典-linux programming simple example, covering file I/O, multi-threaded, multi-process, socket, such as Introduction to Classical
- 2022-08-15 23:22:21下载
- 积分:1
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int show_char(int n, const char *name, chtype code)
{
const int height =...
int show_char(int n, const char *name, chtype code)
{
const int height = 16
int row = 4 + (n height)
int col = (n / height) * COLS / 2
mvprintw(row, col, " *s : ", COLS/4, name)
addch(code)
return n + 1
}- int show_char(int n, const char*name, chtype code)
{
const int height = 16
int row = 4+ (n height)
int col = (n/height)* COLS/2
mvprintw(row, col, "*s : ", COLS/4, name)
addch(code)
return n+ 1
}
- 2023-01-13 11:50:04下载
- 积分:1