Bash – Why pwd does not accept long options like –physical

bashpwdrhel

When I display the manual for pwd command, it says that long options like --physical are supported

$ man pwd
PWD(1)                           User Commands                          PWD(1)

NAME
       pwd - print name of current/working directory

SYNOPSIS
       pwd [OPTION]...

DESCRIPTION
       Print the full filename of the current working directory.

       -L, --logical
              use PWD from environment, even if it contains symlinks

       -P, --physical
              avoid all symlinks

However, it fails when I type the following

$ pwd --physical
-bash: pwd: --: invalid option
pwd: usage: pwd [-LP]

Why are long options not working for me?

I'm using RHEL 6.4. No alias for pwd is configured. Looks like it's standard pwd:

$ which pwd
/bin/pwd

Best Answer

bash has a built-in command pwd which is what you are using when you simply type pwd into your shell.

To get the pwd as described by the manpage, you need force use of the external command. You can do this by specifying the full path to the executable (/bin/pwd in your case) or by prepending env before the line: env pwd, which starts the env command which can be used to add settings to the environment (but which is not done here) and then env starts the command specified. As env doesn't have a builtin pwd, the "real" /bin/pwd is executed.

The advantage of the builtin pwd in bash is that bash keeps track of the current directory, so getting the value is at zero cost, whereas the external command needs to search up through the filesystem to determine the path, which is much more IO intensive.

Using grep

Why can't you just use the -r switch to grep to recurse the filesystem instead of making use of find? There are 2 additional switches I'd use too, instead of the -n switch.

$ grep -rHn PATTERN <DIR> | cut -d":" -f1-2

Example #1

$ grep -rHn PATH ~/.bashrc | cut -d":" -f1-2
/home/saml/.bashrc:25

Details

-r - recursively search through files + directories
-H - prints the name of the file if it matches (less restrictive than -l) i.e. it works with grep's other switches
-n - display the line number of the match

Example #2

$ grep -rHn PATH ~/.bash* | cut -d":" -f1-2
/home/saml/.bash_profile:10
/home/saml/.bash_profile:12
/home/saml/.bash_profile_askapache:99
/home/saml/.bash_profile_askapache:101
/home/saml/.bash_profile_askapache:118
/home/saml/.bash_profile_askapache:166
/home/saml/.bash_profile_askapache:218
/home/saml/.bash_profile_askapache:250
/home/saml/.bash_profile_askapache:314
/home/saml/.bash_profile_askapache:2317
/home/saml/.bash_profile_askapache:2323
/home/saml/.bashrc:25

Using find

$ find . -exec sh -c 'grep -Hn PATTERN "$@" | cut -d":" -f1-2' {}  +

Example

$ find ~/.bash* -exec sh -c 'grep -Hn PATH "$@" | cut -d":" -f1-2' {}  +
/home/saml/.bash_profile:10
/home/saml/.bash_profile:12
/home/saml/.bash_profile_askapache:99
/home/saml/.bash_profile_askapache:101
/home/saml/.bash_profile_askapache:118
/home/saml/.bash_profile_askapache:166
/home/saml/.bash_profile_askapache:218
/home/saml/.bash_profile_askapache:250
/home/saml/.bash_profile_askapache:314
/home/saml/.bash_profile_askapache:2317
/home/saml/.bash_profile_askapache:2323
/home/saml/.bashrc:25

If you truly want to use find you can do something like this to exec grep upon finding the files using find.

Bash – PS1=’$(pwd)’ why this works and why is this different from PS1=$(pwd)

When you simply assign a value to a variable, the $(...) expression is evaluated unless it is enclosed in single quotes (or backslash-escaped). To understand, try and compare these two:

A=$(pwd)
echo "$A"
B='$(pwd)'
echo "$B"

The value of A immediately becomes the string /home/yourusername and obviously it's not remembered where this string comes from, it stays the same even if you change directory. The value of B, however, becomes the literal string $(pwd) without getting interpreted.

Now, in the value of PS1 something special happens: whenever the prompt is printed, certain special constructs are interpreted, e.g. the command substitution $(...) is performed exactly the way it happened above at the assignment to the A variable. Obviously if your PS1 contains the literal string of your home directory (as above with A) then there's no way it could change. But if it contains the string $(pwd) (as above with B) then it is evaluated whenever the prompt is printed and hence your actual directory is displayed.