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I wanted to do a specific ps run on a Mac OS X 10.6 machine. According to man, the -f flag is available in BSD. Is it not available in Mac OS X? If so, why does man list it as an option?

I get illegal option and it highlights the -f flag. I have to sudo to get it to execute.

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According to man ps:

Display the uid, pid, parent pid, recent CPU usage, process start time, controling tty, elapsed CPU usage, and the associated command.

Works fine on my machine:

$ ps -f
  UID   PID  PPID   C     STIME TTY           TIME CMD
  501 58104 58103   0   0:00.01 ttys000    0:00.02 -bash

When getting illegal option then, according to some James Sadler, you need to run export COMMAND_MODE=unix2003. Indeed, on my account (admin, but not root), it is set like that. My OS X 10.6 is an upgrade going back to Tiger. In the same post, James also claims:

I have found that launching ps via iTerm would exhibit the problem but not when run via

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Bizarre. I get illegal option and it highlights the -f flag. I have to sudo to get it to this normal ? Thanks. – Scott Davies Oct 30 '10 at 20:25
@Scott which ps? Mine is /bin/ps. – Daniel Beck Oct 30 '10 at 21:04
@Scott, see my edit. – Arjan Oct 30 '10 at 21:08
...though, after running unset COMMAND_MODE on my machine ps -f still runs fine without sudo. Also, sudo bash -c set | grep COMMAND_MODE yields nothing, though sudo ps -f runs fine too. Are you using bash and Terminal, @Scott? – Arjan Oct 30 '10 at 21:25
Hi Arjan! I am using bash and a 3rd party console app. – Scott Davies Nov 24 '10 at 0:58

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