Why does cd print when run in command substitution?

If I use the 'cd' BASH built-in in a command substitution, it prints extra stuff to stdout, but only when piped to, eg., less.

$ echo `cd .`

# The output is a single newline, appended by echo.


$ echo `cd .` | less
# less displays:
ESC]2;my.hostname.com - tmp/testenv^G
(END)

What's going on there? This behavior isn't documented in the bash man page for cd. Obviously, running just 'cd' in a command substitution is silly, but something like

NEWDIR=`cd mypath; pwd`

could be useful.

I solved this by instead using

NEWVAR=`cd mypath > /dev/null 2>&1; pwd`

but I still want to know what's going on.

Distro: Scientific Linux SL release 5.5 (Boron)

reasgt

Posted 2012-08-29T19:39:07.177

Reputation: 153

I cannot replicate the described behaviour. Do you use $CDPATH? It makes cd output the real path. Or have you defined an alias for cd? – choroba – 2012-08-29T19:44:05.700

Run type cd, result should be cd is a shell builtin. Make sure echo \builtin cd .` | less` results in the same. It forces use of the shell builtin command. Cannot repro, bash 3.2.48(1) on OS X. – Daniel Beck – 2012-08-29T19:47:37.633

I think something must have changed since bash 3.2. See this thread: http://lists.gnu.org/archive/html/bug-bash/2007-04/msg00019.html The problem is that POSIX requires printing of the directory when CDPATH was used, even when not interactive. Why this changed in bash 4.x I don't know, but I can't reproduce it there either.

– reasgt – 2012-08-29T19:49:18.673

Use of command circumvents use of the bash builtin, but instead executes the program returned from which cd. – Daniel Beck – 2012-08-29T19:52:49.227

Why does cd print when run in command substitution?

Answers