How can I increment a value in a bash script array by 1?

Question

I'm trying to increment a value in an array by 1 using the following code, however I'm having some problems with it. Please can someone help me out?

myArray[$position]=((${myArray[$position]}++))

Iodism

What's a unix array? Are you scripting here or something? – Shellashellac

That correct - I'm writing a bash script – Iodism

Answer 1

Try this

 myArr[3]=7
 (( myArr[3]++ ))
 echo ${myArr[3]}

 # output
 8

The (( .... )) can perform bash/ksh's math operations, and the variables referenced inside, don't need to be passed out as in your example, you're probably thinking of a similar construct var=$(( ... MathStuff ...)) OR var=$( ... stringStuff ... ) (note the '$' before the opening paren).

Also note that inside (( ... )) you don't need to use the leading '$' for any math variables like $pct or $counter. If you're using arguments to the script or a function like $1, $2, ... $N, THEN you need to use the $, so the value of $1 is used, instead of just '1'. Thanks to @ChrisDown for the reminder!

I hope this helps.

Vasili

Not true, there are some times where you will have to refer to them with a leading $ to force the context ($1, $2 ... $N). – Outherod

Great, that would certainly affect someones results. I'll update when I get back. Thanks for the improvement! – Vasili

Note that the exit status will be non-zero if myArr[3] is 0 before updating. – Chatter

You can also use let myArr[3]++. – Chatter

Answer 2

Increment and update:

array[1]=$((array[1]++))

Atrocity

Arithmetic expansion obviates the need to reassign the element: a=(1 2 3); ((a[0]++)); echo ${a[@]}. – Medley