Bash Process Substitution does not work as 'root' on OS X 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?)

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: