[SOLVED] [bash] dynamic variables

hashbang#! · 01-05-2013, 11:42 AM

I have a lot of repetitive code here and I tried to replace this piece of code with dynamic variables using eval:

Code:

[[ ! -z "$bg" ]] && bg_canvas=$(canvas $bg_colour)
[[ ! -z "$fg" ]] && fg_canvas=$(canvas $fg_colour)
...

I would like to achieve something like this:

Code:

for t in fg bg; do 
    eval [[ ! -z "'$'${t}" ]] && ${t}_canvas=$(canvas '$'${t}_colour)
done

I ran into two problems:

I didn't know how to test the value of dynamic variable '$'${t}
The output of $(canvas $fg_colour) contains spaces, e.g. "#ff0 #00a", which leads to error bash: #00a: command not found

Any help would be very much appreciated.

grail · 01-05-2013, 12:19 PM

Generally there are very few reasons to use dynamic variable names. My suggestion would be to use associative arrays using the appropriate names, bg & fg, and test and set as needed.
This will also negate the issue you are having with spaces as you simply quote the whole variable.

And of course the use of eval should always be kept to a minimum or avoided where possible.

hashbang#! · 01-05-2013, 05:32 PM

Thank you for your response, grail.

On this occasion I would find it more convenient to use dynamic variables as I have already used this technique elsewhere in the script.

My main stumbling block is 2) - assigning values with spaces to dynamic variables. Is there any way round the problem?
For 1) I have a work-around.

grail · 01-06-2013, 12:23 AM

Quote:

On this occasion I would find it more convenient to use dynamic variables as I have already used this technique elsewhere in the script.

As the old saying goes, continuing with a bad idea because you started with it is hardly a reason to continue.

That being said, the only way I know of nullifying the space issue is either:

1. use double quotes where necessary

2. assuming the above does not work, replace all spaces with something like an underscore and then reverse the process when needing the space version

Generally the idea for me would be, if you need to use option 2 then the overhead associated would mean you need to rethink the design choice.

If David H sees this thread he may have other ideas.

millgates · 01-06-2013, 06:14 AM

First of all, don't use eval.
Also, and I can't stress this enough, you shouldn't use eval.
Last, but not least, DON'T USE EVAL!

This is exactly what the associative arrays are for:

Code:

#!/bin/bash

declare -A array
array[bg_color]="foo  bar"
array[fg_color]="baz"

for i in bg fg; do
    echo "i: ${array["${i}_color"]}"
done

Another way of indirection, if you really don't want to use arrays:

Code:

#!/bin/bash

bg_color="foo  bar"
fg_color=baz
for i in bg fg; do
    var="${i}_color"
    echo "$var: ${!var}"
done

If you really really really can't sleep well without using eval, (did I mention you shouldn't?), you have to quote what you don't want to be word-split. Eval will make the line be processed twice, so you may need to use two layers of quotes or escape them:

Code:

#!/bin/bash

bg_color="foo  bar"
fg_color=baz
for i in bg fg; do
    eval echo "$i: \"\$${i}_color\""
done

rknichols · 01-06-2013, 12:51 PM

Quote:

Originally Posted by hashbang#!

I would like to achieve something like this:

Code:

for t in fg bg; do 
    eval [[ ! -z "'$'${t}" ]] && ${t}_canvas=$(canvas '$'${t}_colour)
done

This should work, using an intermediate variable for the parameter name, '!' for indirection, and "declare" to allow expansion of the LHS of the assignment:

Code:

for T in fg bg; do
    Param=${t}_colour
    [[ ! -z "${!t}" ]] && declare ${t}_canvas="$(canvas ${!Param})"
done

hashbang#! · 01-06-2013, 01:01 PM

Quote:

Originally Posted by millgates

First of all, don't use eval.
Also, and I can't stress this enough, you shouldn't use eval.
Last, but not least, DON'T USE EVAL!

You made me smile on this miserable rainy Sunday.

But neither of you mentioned what's so evil about eval.

Quote:

Originally Posted by millgates

Another way of indirection, if you really don't want to use arrays:
[...]

I like the ${!var} style indirection a lot but it doesn't seem to be possible to assign a value to a dynamically referenced variable:

Code:

${!var}=blue

hashbang#! · 01-06-2013, 01:07 PM

rknichols, our posts crossed. I am in the process of testing your suggestion.

hashbang#! · 01-06-2013, 01:19 PM

Quote:

Originally Posted by rknichols

This should work, using an intermediate variable for the parameter name, '!' for indirection, and "declare" to allow expansion of the LHS of the assignment:

Code:

for T in fg bg; do
    Param=${t}_colour
    [[ ! -z "${!t}" ]] && declare ${t}_canvas="$(canvas ${!Param})"
done

That is ingenious!

colucix · 01-06-2013, 02:34 PM

Just an aside note: instead of

Code:

! -z

you can use

Code:

-n

which is the same but more concise.

colucix · 01-06-2013, 02:37 PM

Quote:

Originally Posted by hashbang#!

But neither of you mentioned what's so evil about eval.

Here is a useful link: http://mywiki.wooledge.org/BashFAQ/048 and here another article about associative arrays and indirect reference: http://mywiki.wooledge.org/BashFAQ/006.

hashbang#! · 01-06-2013, 04:13 PM

colucix - thanks for the useful links