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Kris
Guest
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 Posted: Tue May 15, 2007 1:46 pm Post subject: gethostbyaddr() does not return any info |
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Hi,
I'm attempting to obtain the NetBIOS name of given IP address (which is
a NAS device in my LAN).
However, the gethostbyaddr() call always returns NULL and the
WSAGetLastError() returns 6 (=WSA_INVALID_HANDLE).
Does anybody know what I'm doing wrong ? The code below shows following
output:
Address=1711280300
RemoteHost=00000000
Error=6
struct hostent* RemoteHost;
unsigned int Address;
char host_name[]="172.16.0.102";
Address=inet_addr(host_name);
DEBUG("Address=%d",Address);
RemoteHost=gethostbyaddr((char*)&Address,4,AF_INET);
DEBUG("RemoteHost=%08X",RemoteHost);
int Error=WSAGetLastError();
DEBUG("Error=%d",Error);
Thanks,
Kris |
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Hans Galema
Guest
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| Posted: Tue May 15, 2007 2:35 pm Post subject: Re: gethostbyaddr() does not return any info |
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Kris wrote:
| Quote: | Does anybody know what I'm doing wrong ?
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You forgot to initialise.
The Windows Sockets WSAStartup function initiates use of the Windows Sockets DLL by a process
| Quote: | char host_name[]="172.16.0.102";
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How terrible to call an ip-address a hostname!
Add some code before and afther what you already have:
WORD wVersionRequested =MAKEWORD( 1,1);
WSADATA wsaData;
long res=WSAStartup( wVersionRequested, &wsaData );
if(res)
{
ShowMessage ( "res so return;" );
return;
}
char ipaddr [] = "172.16.0.102";
unsigned long inaddr = inet_addr ( ipaddr );
hostent *myhostent = gethostbyaddr( (const char FAR *)&inaddr, sizeof ( in_addr ), 0 );
if ( myhostent ) ShowMessage ( ipaddr + AnsiString ( "\n\n" ) + myhostent->h_name );
else
ShowMessage ( ipaddr + AnsiString ( "\n\n" ) + "no myhostent" );
WSACleanup();
Hans. |
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Kris
Guest
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| Posted: Tue May 15, 2007 3:44 pm Post subject: Re: gethostbyaddr() does not return any info |
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Hans Galema wrote:
| Quote: | Add some code before and afther what you already have:
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Cool. After I changed one little thing in your example code I got it
working perfectly now. I changed the
gethostbyaddr(..., 0 );
into
gethostbyaddr(..., AF_INET );
Thank you very much for your help.
Regards,
Kris |
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