What the heck is ${RA[2]} in this context?

[CrazedNerd]

for person in $ME $3 Pat ${RA[2]} Sue; do

echo $person

done

Overall a pretty simple loop in bash, if i make my name that variable (ME=CrazedNerd), it will loop my name until Sue. However, ${RA[2]} seems completely random and doesn't have any functionality so far. When i run the script as described, it loops my name until sue, and bash gives me no bugs/errors. I tried assigning something to "RA" but that doesn't change it either. All i know is that RA[2] is getting grouped together by brackets.

 

[tymerahir]

RA is an array.
${RA[2]} is the third element of this array.

Run this example:

#!/bin/bash

ME="aaa"
RA=( bbb ccc ddd )

for person in $ME $3 Pat ${RA[2]} Sue; do
echo $person
done

 

[gvisoc]

The braces ${...} are a way to delimit the name of the variable for the intepreter, in context where we want to explicitly indicate what to interpret as the parameter or variable expand.

Try this:

#!/bin/bash
name="This is my name"
name_="This is not my name"

echo "$name"
echo "$name_"
echo "${name}_"

The author of the script wanted the interpreter to evaluate RA[2] and not RA, and then attach [2],

 

[CrazedNerd]

Oh okay, that makes sense, thanks everyone

 

[CrazedNerd]

Just to potentially help someone who's a bash novice like me, here's a full demonstration of a loop and array.

Script syntax:

ME=CrazedNerd
RA=( GOO-GOO GAA-GAA SIM-SIM )

for person in $ME Pat ${RA[0]} ${RA[1]} ${RA[2]} Sue; do
echo $person
done

Command line entry:

ME=CrazedNerd ; RA=( GOO-GOO GAA-GAA SIM-SIM ) ; for person in $ME Pat ${RA[0]} ${RA[1]} ${RA[2]} Sue ; do echo $person ; done

output:

CrazedNerd
Pat
GOO-GOO
GAA-GAA
SIM-SIM
Sue