Bash is slow to start because of this line in .bashrc. What could cause this?

My .bashrc file contains a line to this effect:

alias prog="/path/to/script.sh $(find $(pwd) -name prog)"

When I comment out this line, Bash starts almost instantly when I open a new terminal. With this line, there is a 4-5 second delay before my cursor shows up.

Removing the nested commands $(pwd), etc. speeds it up again as well. Why is this happening? Can I still use nested commands somehow?

BBales

Posted 2017-05-14T21:57:47.447

Reputation: 293

4The problem is the find command, which is execute at boot time and can take a long time to execute, depending on the directory structure. If you use single- instead of double-quotes, find will be executed at run time, when the alias is invoked. It depends on which working directory you want to use, that current at declaration time, or at run time. By the way $(pwd) is a rather inefficient way to express . or $PWD. – AFH – 2017-05-14T22:14:40.803

I know that's not what you are asking, but this looks a lot like a XY problem. Why do you need to locate the script using find? Once you installed it, you know where it is. Even if you update it frequently and change its name, you can maintain a symbolic link to it so it's always available under a fixed name.

– sleske – 2017-05-15T07:41:28.270

5@AFH Please post answers as answers, not comments. – David Richerby – 2017-05-15T11:31:28.613

1@DavidRicherby - I wanted to make sure that this was indeed the answer before posting. I was puzzled that the script would work on files found within an arbitrary directory. – AFH – 2017-05-15T14:13:18.817

Answers

Because the command substitution is inside double-quotes, it is evaluated at the time that the command is defined. This causes find to look through your hard disk contents while .bashrc is running.

You, by contrast, appear to want it evaluated at the time of use. In that case, use single quotes:

alias prog='/path/to/script.sh $(find "$(pwd)" -name prog)'

Note that this alias will fail if any of the files found have whitespace in their names. To avoid that:

alias prog='find . -name prog -exec /path/to/script.sh {} +'

This latter form will work reliably for any kind of file name.

John1024

Posted 2017-05-14T21:57:47.447

Reputation: 13 893

1'time of use'. And your second line avoids spawning a subshell. – rleir – 2017-05-15T03:57:44.347

11On the downside, now every time you type prog, find would need to be run, whereas before, it would be run only once (on shell startup). – el.pescado – 2017-05-15T06:21:43.273

Then again, owing to file caching, the second time you run the same find it's usually much faster. – reinierpost – 2017-05-15T08:53:31.640

21@el.pescado On the upside, it's hard to imagine that the intended behaviour is to run the command on the files that existed at the time the user logged in, rather than the files that exist when they executed the command to process them. – David Richerby – 2017-05-15T11:33:10.193

2@DavidRicherby Maybe we can cache the path of prog and re-run find if the old path no longer exists? – Alex Vong – 2017-05-15T20:17:33.050