[Solved]Bash:Subtract a set of numbers from each other in pairs?

Question

Solution: see my reply.
I know the description isn't exact, there really isn't an easy way to describe it with the limits of the title so the circumstances are these:
say I have arrays which contain small numbers like so:
ex. 12, 7 and 13
I which is greater than the other when comparing the following number in sequence and need to subtract the lower one from it. the 3 numbers above in that order will end up producing two simple problems:
(12 - 7) and (13 - 7). The results will be stored in an array in the order they were performed (left to right)
Currently I'm trying to accomplish this with a while loop and some if statements
Note: the array "Tree" is named such so I can print the results somewhat like so after performing this operation until only one result is left:
12   7   13
     5   6
        1
#!/usr/bin/env bash
Num_Set=( 12 7 13 )
A=0 ; B=1
while (( ${#Num_Set[B]} >= 1 )) ; do
if (( ${Num_Set[A]} >= ${Num_Set[B]} )) ; then
Tree[A]=$(( ${Num_Set[A]} - ${Num_Set[B]} ))
elif (( ${Num_Set[A]} < ${Num_Set[B]} )) ; then
Tree[A]=$(( ${Num_Set[B]} - ${Num_Set[A]} ))
fi
A=$(( A++ )) ; B=$(( B++ ))
done
echo "${Num_Set[@]}"
echo "${Tree[@]}"
I believe i'm on the right track but this becomes an endless loop for reasons unknown for to me.
Any bash hackers out there know whats happening here?
Last edited by Thme (2013-06-12 18:57:40)

Answer 1

And bash has associative arrays as well. Your claim that the $ is not recommended is not valid at all:
declare -A aa
aa[bb]=100
aa[0]=1
bb=0
echo $(( aa[bb] + 1 ))
101
echo $(( aa[$bb] + 1 ))
2
My apologies however I tested my original post with http://www.shellcheck.net/ upon finding the site through an unrelated thread. This was an example test to illustrate what was commented on and what was not. This works in the shell:
A=( 2 1 10 )
B=2
echo $(( A[0] + A[B] - ${A[${A[$B]}]} - A[$B] + ${A[A[1]]} + A[0] - ${#A[2]} ))
23
However ${A[${A[$B]}]} and ${A[A[1]]} were commented on by shellcheck with "don't use $ on variables in (( ))". Aside from that it does still work. Also I tested this as well:
echo $(( A[${A[0]}] + 1 ))
And although it works in my bash it gets "This $((..)) expression is malformed." and "Unexpected "}". Fix any mentioned problems and try again." when tested in shellcheck. I suspect that maybe a fault on shellcheck but either way it looks less readable than just A[A[0]]. So what I've gathered from all this is that, although the above examples all work, the following ones below are probably the best practice here for use in (( )).
#!/usr/bin/env bash
declare -A AA
AA[BB]=3
AA[0]=1
BB=0
A=( 2 1 10 )
B=1
echo $(( A[0] + 1 ))
echo $(( B + 1 ))
echo $(( A[0] + 1 ))
echo $(( A[B] + 1 ))
echo $(( A[A[0]] + 1 ))
echo $(( ${#A[2]} + 1 ))
echo $(( ${A[2]/0/1} + 1 ))
echo $(( A[BB] + 1 ))
echo $(( AA[$BB] + 1 ))
echo $(( AA[BB]++ ))
So in the cases of using arrays in manners like ${#A[2]}, ${A[2]/0/1} and AA[$BB] ( AA[BB] being different. ) using $ would be necessary inside of (( )). If I missed any or there are any needed corrections please note them as I'm concerned with good practices here. I'll leave it at that. Again thanks for the help everyone.
Last edited by Thme (2013-06-15 00:49:54)