Defining an array from concatenated variables in bash

Question

Outside of the code I posted below I defined an array fanrp0[1] and fanrp0[2] I did the same for fans 0 through 3. I'm trying to make fstat change and echo each of the fans in two arrays.

    for (( f=1; f<=2; f++ ))
     do
       for (( f2=0; f2<=3; f2++ ))
        do
          fstat="${(fanrp$f2)[$f]}"
          echo $fstat
        done
     done

It looks like you're trying to do what bash refers to as an indirect reference, which are apparently fairly difficult and cumbersome to do with array variables, judging by [this Bash FAQ page](http://mywiki.wooledge.org/BashFAQ/006), which I found on [this question](http://stackoverflow.com/questions/11180714/how-to-iterate-over-an-array-using-indirect-reference). You can accomplish what you need with `eval` *somewhat* more easily, as in `eval "fstat=\${fanrp$f2[$f]}"`, but keep in mind that there are some security issues with using `eval`. I'd start by reading that Bash FAQ page I linked to. — Mike Holt, Mar 14 '14 at 00:23
Indirect array references are easier once you start thinking of the `[...]` not as indexing syntax, but as part of the name of the variable. If you have an array `foo`, then `foo[0]`, `foo[1]`, etc. are just oddly named variables, each with their own value. Then `x=foo[0]` leads naturally to `${!x}` being equivalent to `${foo[0]}`. — chepner, Mar 14 '14 at 02:51

score 3 · Accepted Answer · answered Mar 14 '14 at 02:56

3

If fanrp0, fanrp1, fanrp2, and fanrp3 are all arrays, you can write your loops as

for (( f=1; f<=2; f++ ))
do
  for (( f2=0; f2<=3; f2++ ))
  do
    x="fanrp$f2[$f]"
    fstat="${!x}"
    echo $fstat
  done
done

Since f2 is used only to generate the array names, a slightly cleaner syntax would be

for f in {1..2}; do
  for arr in fanrp{0..3}; do
    x="$arr[$f]"
    fstat=${!x}
    echo $fstat
  done
done

answered Mar 14 '14 at 02:56

chepner

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The first option actually gave me a bad substitute error, that's what I wrote for my first attempt and I couldn't figure out why I was getting that error. However, I think your second solution made a lot more sense to me, thanks it worked great I wasn't aware of the for arr – Cfoote7 Mar 14 '14 at 04:57

Defining an array from concatenated variables in bash

1 Answers1