var=$(uname -r) or var=`uname -r` ?

dtw · 2005-06-20 08:20:39

Are they the same? Do they pose the same security risks in a PKGBUILD?

iphitus · 2005-06-20 09:48:03

I always used the latter one.... but i think they're both two ways of the same thing...

T-Dawg · 2005-06-20 10:04:49

ditto what iphitus said. I'm not sure how you mean with "security risk" though, what are you trying to do with it?

dtw · 2005-06-20 10:51:55

if i put `rm -rf /` in it for example and the pkgbuild is parsed...

sudman1 · 2005-06-20 11:12:44

They are the same, but IIRC $(...) is the preferred method

juergen · 2005-06-20 14:27:39

sudman1 wrote:

They are the same, but IIRC $(...) is the preferred method

Not exactly the same. Try:

arg=foo
echo `echo $arg`
echo $(echo $arg)

This will yield:

foo
$arg

There is no difference when not using backslash escape character. Best check GNU Info manual.

Jürgen

Dusty · 2005-06-20 14:39:34

The $() syntax is bash, the backtick syntax is older sh, though of course bash supports it. THe reason bash included the $() syntax is that its easier to nest. So use $() when possible but if you need to deal with sh, use backticks.

Dusty

kpiche · 2005-06-20 16:58:37

I understand that $() is preferred for bash. Use:

arg=foo
echo `echo $arg`
echo $(echo $arg)

Arch Linux

#1 2005-06-20 08:20:39

var=$(uname -r) or var=`uname -r` ?

#2 2005-06-20 09:48:03

Re: var=$(uname -r) or var=`uname -r` ?

#3 2005-06-20 10:04:49

Re: var=$(uname -r) or var=`uname -r` ?

#4 2005-06-20 10:51:55

Re: var=$(uname -r) or var=`uname -r` ?

#5 2005-06-20 11:12:44

Re: var=$(uname -r) or var=`uname -r` ?

#6 2005-06-20 14:27:39

Re: var=$(uname -r) or var=`uname -r` ?

#7 2005-06-20 14:39:34

Re: var=$(uname -r) or var=`uname -r` ?

#8 2005-06-20 16:58:37

Re: var=$(uname -r) or var=`uname -r` ?

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