accessing loop variable in a loop [duplicate]

Your problem is that $sum_$x is interpreted as the concatenation of $sum_ and $x:

~$ for x in $LIST1;  do   echo $sum_$x;  done
1
2
3
4

Hence your problem. What you want is using

~$ for x in $LIST1;  do   echo $((sum_$x));  done
15665
20981
24995
23308

I don't really understand what your script is supposed to be doing and you haven't told us what language you are writing in. I will therefore assume bash and that you are aware of the fact that the values you have in your question don't make your script do anything very useful.

In general, concatenating strings to get a variable name is both quite cumbersome and unneeded. Just use arrays instead:

#!/usr/bin/env bash
load=(15231 20547 24561 22874)
f4=434

for((i=0;i<${#load[@]};i++))
do
    sums[i]=$((${load[i]} + $f4))
done

min=$((${sums[0]}))
for((i=1;i<${#load[@]};i++))
do
    if [ "${sums[i]}" -lt "$min" ]
    then
       min="${sums[i]}"
       f3="$i"
    fi
done

printf "%s\n%s" "$min" "$f3"

1. a more bash way for what you need:

load_1=15231  
load_2=20547  
load_3=24561
load_4=22874
f4=434
line=`(set -o posix ; set) | grep load_  | sort -t = -n -k2 | head -1`
min=`echo $line | cut -d= -f2`
echo $((min+f4))
echo $line | cut -d= -f1 | cut -d_ -f2

This results in: 15665 1

if you want to use your code anyway:

2. you can avoid using LIST1 by for x in {1..4}

3. set f3=1 when you set min, otherwise - as in your example - it isn't set

load_1=15231
load_2=20547
load_3=24561
load_4=22874
f4=434
min=0
f3=1

for n in 1 2 3 4
do   [ "$((sum_$n=load_$n+f4))" -lt \
       "$((min=min?min:sum_$n))" ] &&
     min=$((sum_$n)) f3=$n
     printf '$%s =\t%d\n' \
            "sum_$n" "$((sum_$n))" \
             min "$min" f3 "$f3"
done

The math expansion will expand the shell variables first. When it does it will arrive at a statement that looks like:

sum_[1234] = load_[1234] + f4

...where the brackets represent each iteration in the loop, and which is valid arithmetic syntax because strings represent variable names for integers in a shell arithmetic expansion. It will set those variables in the current shell context even when it is used as an argument to [ test ] - so $sum_1 is simultaneously set to the value 15231 and compared to $min which is 0.

But based on your values $min is never adjusted according to $sum_$n because no value for $sum_$n is ever less than $min.

OUTPUT

$sum_1 =    15665
$min =  15665
$f3 =   1
$sum_2 =    20981
$min =  15665
$f3 =   1
$sum_3 =    24995
$min =  15665
$f3 =   1
$sum_4 =    23308
$min =  15665
$f3 =   1

In fact, we could put the entire thing in a single math expansion and leave the test out entirely:

load_1=15231
load_2=20547
load_3=24561
load_4=22874
f4=434
f3=$((!(min=0)))
for n in 1 2 3 4
do  echo \$min = "$((
           min = ( min    >  ( sum_$n = load_$n+f4 ) )
               ? ( sum_$n + !( f3     = $n         ) )
               : ( min    ?    min    : sum_$n     )
))"; done

That sets all variables appropriately for every iteration, but only expands to $min's value each time. It prints:

$min = 15665
$min = 15665
$min = 15665
$min = 15665