312-50V11 · Question #313
What is the outcome of the comm"nc -l -p 2222 | nc 10.1.0.43 1234"?
The correct answer is B. Netcat will listen on port 2222 and output anything received to a remote connection on 10.1.0.43. The piped netcat command creates a relay: the first nc listens on port 2222 and forwards all received data to a remote host via the second nc instance.
Question
What is the outcome of the comm"nc -l -p 2222 | nc 10.1.0.43 1234"?
Options
- ANetcat will listen on the 10.1.0.43 interface for 1234 seconds on port 2222.
- BNetcat will listen on port 2222 and output anything received to a remote connection on 10.1.0.43
- CNetcat will listen for a connection from 10.1.0.43 on port 1234 and output anything received to
- DNetcat will listen on port 2222 and then output anything received to local interface 10.1.0.43.
How the community answered
(32 responses)- A13% (4)
- B78% (25)
- C6% (2)
- D3% (1)
Why each option
The piped netcat command creates a relay: the first nc listens on port 2222 and forwards all received data to a remote host via the second nc instance.
Netcat does not use an IP address as a listening interface in this syntax, and '1234' is a port number not a duration in seconds.
In this command, 'nc -l -p 2222' listens on port 2222 for incoming connections, and the pipe operator passes any received data as stdin to 'nc 10.1.0.43 1234', which opens an outbound connection to host 10.1.0.43 on port 1234. This creates a one-way data relay from the listening port to the remote destination, commonly used in port forwarding and pivoting scenarios.
The '-l -p 2222' flags specify a listening port of 2222, not a connection from a specific remote IP; the second nc argument specifies the destination, not a source filter.
10.1.0.43 is the destination of the second nc instance (a remote host), not a local interface; the data is forwarded outbound to that remote address.
Concept tested: Netcat pipe relay for port forwarding
Source: https://nmap.org/ncat/guide/ncat-usage.html
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