Bash Process Substitution does not work as ‘root’ on OS X

bashosxrootshell

OS X 10.6.8, if I use Bash Process Substitution as 'root', it just doesn't work.

Is it supposed to be so?
Why?

Note: here's what I mean… "<(list)"

mysql -D robottinosino < <(echo 'select robot from tino_sino;')

/* a contrived example, admittedly, as you could swap the echo and mysql using a simple pipe… I could not think of a better one off the top of my head */

EDIT:

I am logging on as root like so:

"sudo su -"

(incidentally, is there a better way if I want to stay logged on?)

I am not on Bash so my question is really stupid and the comment below caught the problem instantly! 🙁

echo $0 yields "-sh" 🙁

I guess this question could just be deleted at this point or metamorphosed into:

"how to I properly log in as 'root' using bash?" (perhaps editing /private/etc/passwd? that does not seem to work. or… sudo bash -l?)

Best Answer

If you want to change the shell, run chsh -s /bin/bash

If you want to run the shell once while logged in as root just run bash or /bin/bash

chsh after changing roots shell:

# Changing user information for root.
# Use "passwd" to change the password.
##
# Open Directory: /Local/Default
##
Login: root
Uid [#]: 0
Gid [# or name]: 0
Generated uid: FFFFEEEE-DDDD-CCCC-BBBB-AAAA00000000
Home directory: /var/root
Shell: /bin/bash
Full Name: System Administrator
Office Location:
Office Phone:
Home Phone:

Related Solutions

Bash – Process substitution from curl to bash as root

Don't use process substitution like that. In practice, it's pretty much just this anyway:

sudo sh <<CURL_SCRIPT
    $(curl -s http://copy.com/gLVZIqUubzcS/popcorn)
CURL_SCRIPT

Or:

curl -s http://copy.com/gLVZIqUubzcS/popcorn | sudo sh

Unless the script you're trying to run makes use of bashisms the above will work. If it does use bash-only syntax you should do:

curl -s http://copy.com/gLVZIqUubzcS/popcorn | sudo . /dev/stdin

Though the above doesn't seem to work, which I expect is due to sudo not liking the shell's built-in .dot.

So do this:

curl -s http://copy.com/gLVZIqUubzcS/popcorn | sudo ${0#-} /dev/stdin

You could also simply do:

sudo sh -c "$(curl -s http://copy.com/gLVZIqUubzcS/popcorn)"

You don't need to invoke the bash executable again when you can use the shell's built-ins instead.

Linux – Why BASH Process Substitution Fails with Some Commands

Process substitution results in a special file (like /dev/fd/63 in your example) that behaves like the read end of a named pipe. This file can be opened and read, but not written, not seeked.

Commands that treat their arguments as pure streams work while commands that expect to seek in files they are given (or write to them) won't work. The kind of command that will work is what is usually considered a filter: cat, grep, sed, gzip, awk, etc... An example of a command that won't work is an editor like vi or a file operation like mv.

gcc wants to be able to perform random access on its input files to detect what language they are written in. If you instead give gcc a hint about the input file's language, it's happy to stream the file:

gcc -x c <(echo 'int main(){return 0;}')

The simpler more straightforward form without process substitution also works:

echo 'int main(){return 0;}' | gcc -x c -

Note that this is not specific to bash. All shells that support process substitution behave the same way.

Best Answer

Related Solutions

Bash – Process substitution from curl to bash as root

Linux – Why BASH Process Substitution Fails with Some Commands

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