See example http://www.huihoo.org/ace_tao/ACE-5.2+TAO-1.2/ACE_wrappers/docs/tutorials/Chap_7/ex01.html[^]

try using _heapwalk

Why is it so difficult to find project menu ??? Alternatively Press Alt + F7

"I want to know how this is done?" Do any of below link answer your question http://en.wikipedia.org/wiki/Domain_name_system[^] http://stackoverflow.com/questions/307291/how-does-the-google-did-you-mean-algorithm-work[^] other answer i do not know

Question not clear. Please Try explaining again

which application is it. check whether it has any command line interface. through which you can pass the username and password.

How about including Ping in your appication... http://www.codeproject.com/KB/IP/Ping_and_Traceroute.aspx[^] Can easily retrieve the values from here.

Are you looking to develop plugin's that you can attach to these messangers. If yes Check the below links, on which platform these are developed should not matter. finally you just have the executable. http://developer.yahoo.com/messenger/[^] http://dev.aol.com/aim/plugins[^] http://channel9.msdn.com/forums/TechOff/101812-MSN-Messenger-Development/[^] http://forum.skype.com/index.php?s=606f036c3cec316698882051231177d0&showforum=16[^] http://code.google.com/apis/apps/[^]

I think you need to take backup of all files and reinstall your OS. Or clean your system. Remove unwanted application etc check this link http://www.microsoft.com/windowsxp/using/setup/expert/northrup_restoreperf.mspx[^]

Please Explain what exactly are you trying to do. Did you analyse exactly what was shown by wireshark. Please post it so we know more info. My guess is you icmp message was 12 byte but the whole packet encapsulated by lower layers(udp and ethernet frame) made it to look of wrong size to you but is of correct size. the icmp reply also contain the orginal ping request which is again of different size. Did you send 12 bytes and wireshark showed 150 byte. or you sent 150 bytes and wireshark showed 12 bytes. Just check what data are you giving to the socket????

I think in linux ioctl(socketid, SIOCGIFCONF, &ifc); can be used. I think you can also use function pcap_findalldevs in libpcap library(opensource). alternatively you can also see the linux specific implementation of pcap_findalldevs in libpcap.

I just meant some other hotfix might be able to solve the problem...any way good luck this is all I know

okay you can take any other example....this was just to prove the point....

he never says the sample size is 6 he just gave you 6 to you because he cannot go on posting all samples

I had a similar problem in which the standard MFC application was crashing on certain action inside file open dialog on few machines with a simple code..... The problem for me was solved for me after downloading a windows hotfix refer to this link if you can find something http://www.bokebb.com/dev/english/1960/posts/196055272.shtml[^]

The problem is solved after patching the OS with microsoft hotfixes

But he has just given you few samples of input and output of a function. all the remaining output cannot be assumed to be zero what you are saying is this: if i say f(100) = 10 f(81) = 9 f(64) = 8 f(49) = 7 f(36) = 6 f(25) = 5 here its a simple example in which output is just square root of input now based on your assumption f(n) = 0, does that mean square root of all remaining numbers are zero

This question reminds me of the hollywood movie "Beautiful Mind".

why is it assumed that f(n) = 0 for all other n

This problem can be seen as finding intersection of two circle. also this problem will have two solutions. How to imagine it as problem of intersection of two circles. Circle 1 center point be A with radius AC and circle 2 as of center B and radius BC. Note: ACB angle 90 degree information is redundant and is not required check out the below link to find the solution of intersection points of two circles and well as the c code to do it. The c code can be directly used for your purpose. http://ozviz.wasp.uwa.edu.au/~pbourke/geometry/2circle/[^] You could have found a faster solution in mathematics and algorithm discussion section.