Re: Can't get remote host and port after CFStreamCreatePairWithSocketToNetService
Re: Can't get remote host and port after CFStreamCreatePairWithSocketToNetService
- Subject: Re: Can't get remote host and port after CFStreamCreatePairWithSocketToNetService
- From: Josh Graessley <email@hidden>
- Date: Mon, 13 Apr 2015 13:44:28 -0700
> On Apr 13, 2015, at 1:25 PM, Jens Alfke <email@hidden> wrote:
>
>
>> On Apr 13, 2015, at 10:12 AM, Jeff Johnson <email@hidden> 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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