可视电话程序源码 可视电话程序源码

一个TCP聊天例子

本人收集的一个windwos下的端口扫描程序源代码．-I collected a windwos the port scanning source code.

MYIE流览器源代码(VC++),vb环境运行 ============= For your information: I build myie32src with 1) Windows2000Professional + sp2 2) VC6.0 + sp5 + new "tlogstg.h" + new "exdisp.h". Copy "tlogstg.h" and "exdisp.h" to your VC include directory, like "C:Program FilesMicrosoft Visual StudioVC98Include". Please backup and then overwrite any old file.-MYIE browser source code (VC), vb environment running ============= For your information : I build with a myie32src) Windows2000Professional sp2 2) VC6.0 SP5 new "tlogstg.h" new "exdisp.h." Copy "tlogstg.h" and "exdisp.h" to your VC include directory, like "C : Program Files Microsoft Visual Studio VC98 Include." Please backup and then overwrite any old file.

一个封装了Winpcap中二层函数的类-a package of Winpcap Layer-2 function category

采用socket编程实现的一个双方发送信息的聊天室程序-Use socket programming to send information on the two sides of a chat room program

可以实现注册，登录，聊天等功能！要用到SQL数据库存储数据！-Can be registered, log in, chat and other functions! SQL database used to store data!

IOCP完成端口编程，源码可以直接使用，学习完成端口的好的素材-IOCP completion port programming, source code can be directly used to complete the port of a good learning material

This the FreeMED fax transmission API, and is used to send faxes using the HylaFax fax system

介绍如何在程序运行时获取当前机器的网卡信息。-procedures on how to run the current machinery access card information.

纯ALOHA协议 在该协议下，用户只要有帧产生，就会传送出去，发生冲突时，就等待一个随机时间后再重发。 所以对当前帧时，只有两种情况下可以发送成功： 1）当前帧时只有一帧要发送，且当前帧时的前一帧时无帧发送； 2）当前帧时只有一帧要发送（发送时刻为t），且前一帧时的最后一帧的发送时间不在（t- FRAMETIME）~ t这段时间内。 其他情况下都会产生冲突。 发生冲突时，错误帧继续传送，用户重新发送帧，设定一个随机数Rt，将该帧添加到第Rt/ FRAMETIME帧时上去，并设该帧的相对时间为Rt% FRAMETIME。这样即相当于用户等待一随机时间后重发。 对每一帧时进行上述控制，最终可得到成功发送的次数Suc，将其除以Suc = Suc /ARRAYLENGTH即得平均每帧时发送成功的帧数。 对于不同G，得到不同FrameArray，所得的Suc也不同，用一个数组来保存G与Suc的对应关系，再将其画出来。 -pure ALOHA agreement under the agreement, as long as users have a frame, it will transmit out, conflict, will wait for a random period of time before re-issued. Therefore, the current frame, there are only two circumstances can this success : 1) the current frame only one to send, and the current frame before one-time frame sent; 2) the current frame only to send a (t this moment), but before a final of a transmission time not (t-FRAMETIME) ~ t this time. Other cases will have conflict. Conflict, to continue to disseminate the wrong frame, users re-sending fram