bash: differing newline count when assigning result to variable [duplicate]

Answer 1

• Command substitution itself runs in a subshell so thats one bash process

• your search for bash ($0) i.e. grep -c bash also ends up in the process table at that time so thats another process (grep) containing string bash. Note that, this might not show up in the process table at the time of running, depending on how busy your system is.

• And you have two (or whatever) actual bash processes (sessions) running presumably are the rest

You can use a Regex trick to get rid of the false positive i.e. grep one from count:

ps ax | grep -c "[b]ash"

It would still count the subshell while doing command substitution:

var=$(ps ax | grep -c "[b]ash")

So you need to manually remove one from this count.

Example:

$ var=$(ps ax | grep -c "bash")    
$ echo $var
4

$ var=$(ps ax | grep -c "[b]ash")   
$ echo $var
3

Answer 2

Your command counts the grep command line too.

ps ax | grep -v grep | grep -c "$0"

should omit the grep from the count

Answer 3

Quoting Siegex:

Because the grep process itself is being returned by ps.

You can either of these:

"trick" grep to not match itself by surrounding one of the search characters in a character class [ ] which doesn't change the functionality:

Or, in this case,

Pipe to grep -v grep, so that the process doesn't match:

var=$(ps ax | grep -v grep | grep "$0")

---

See an example. Here we have a process sleep:

$ sleep 20 &
[1] 5602

If we check for it in the output of ps it appears twice!

$ ps -ef| grep sleep
me   5602  5433  0 09:49 pts/2    00:00:00 sleep 20
me   5607  5433  0 09:49 pts/2    00:00:00 grep --colour=auto sleep

So we can either use a character class:

$ ps -ef| grep [s]leep
me   5602  5433  0 09:49 pts/2    00:00:00 sleep 20

Or grep out the grep process:

$ ps -ef| grep sleep | grep -v grep
me   5602  5433  0 09:49 pts/2    00:00:00 sleep 20