> On Apr 13, 2015, at 1:25 PM, Jens Alfke <[email protected]> wrote:
> 
> 
>> On Apr 13, 2015, at 10:12 AM, Jeff Johnson 
>> <[email protected]> wrote:
>> 
>> I want the host and port is to send a "Host" header. I might be able to get 
>> away without it, but I'd prefer to adhere to standards.
> 
> I ended up having to implement this earlier today; here’s the code I came up 
> with. “_inputStream” is an NSInputStream opened on a TCP socket. Disclaimer: 
> I haven’t tested it much yet, and not at all with IPv6 addresses.
> 
> - (NSString*) remoteHost {
>     // First recover the socket handle from the stream:
>     NSData* handleData = CFBridgingRelease(CFReadStreamCopyProperty(
>                                                   (__bridge 
> CFReadStreamRef)_inputStream,
>                                                   
> kCFStreamPropertySocketNativeHandle));
>     if (!handleData || handleData.length != sizeof(CFSocketNativeHandle))
>         return nil;
>     CFSocketNativeHandle socketHandle = *(const 
> CFSocketNativeHandle*)handleData.bytes;
>     // Get the remote/peer address in binary form:
>     struct sockaddr_in addr;
>     unsigned addrLen = sizeof(addr);
>     if (getpeername(socketHandle, (struct sockaddr*)&addr,&addrLen) < 0)
>         return nil;
>     // Format it in readable (e.g. dotted-quad) form, with the port number:
>     char nameBuf[INET6_ADDRSTRLEN];
>     if (inet_ntop(addr.sin_family, &addr.sin_addr, nameBuf, 
> (socklen_t)sizeof(nameBuf)) == NULL)
>         return nil;
>     return [NSString stringWithFormat: @"%s:%hu", nameBuf, 
> ntohs(addr.sin_port)];
> }

This seems sub-optimal as there can be many hostnames that resolve to a single 
address. A reverse-lookup of the address will yield one of those hostnames, but 
it might not be the one you’re looking for.

Also, what does this do if you connect through a proxy? The native handle could 
be a socket to a SOCKS (or other) proxy, right?

A radar requesting a better way to do this might be a good idea. If you file 
one, please forward me the radar #, thanks.

-josh


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