0

I saw this in a bash script, and trying to make sense of it:

exec {STDIN}>&0
exec {STDOUT}>&1
exec {STDERR}>&2

Are the {STDIN}, {STDOUT} etc "magic strings" recognized by exec directive in bash scripts?

Isn't STDOUT already set to &1 (and STDERR set to &2, ...) anyway?

What do these lines do?

ahmet alp balkan
  • 601
  • 1
  • 8
  • 21

1 Answers1

3

From man bash:

REDIRECTION

Before a command is executed, its input and output may be redirected using a special notation interpreted by the shell. [...]

Each redirection that may be preceded by a file descriptor number may instead be preceded by a word of the form {varname}. In this case, for each redirection operator except >&- and <&-, the shell will allocate a file descriptor greater than or equal to 10 and assign it to varname. If >&- or <&- is preceded by {varname}, the value of varname defines the file descriptor to close.

So the lines in your example:

exec {STDIN}>&0
exec {STDOUT}>&1
exec {STDERR}>&2

have the effect of assigning three new file descriptors that are duplicates of the current descriptors 0, 1 and 2, and assigning the values of those new descriptors to the shell variables $STDIN, $STDOUT and $STDERR respectively. After those commands have been executed you should be able to see the values of the duplicate descriptors:

$ echo $STDIN $STDOUT $STDERR
10 11 12

The most likely reason for doing this is to save the current disposition of stdin, stdout and stderr so that the script can freely redirect descriptors 0, 1 and 2 to various places as needed by some parts of the script, and then restore them to their original targets by doing something like:

exec 0>&$STDIN
exec 1>&$STDOUT
exec 2>&$STDERR
ottomeister
  • 383
  • 1
  • 6