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]}++))
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What's a unix array? Are you scripting here or something?Carl Norum– Carl Norum2011-11-19 16:50:07 +00:00Commented Nov 19, 2011 at 16:50
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That correct - I'm writing a bash scriptuser843337– user8433372011-11-19 16:50:36 +00:00Commented Nov 19, 2011 at 16:50
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2 Answers
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.
4 Comments
Chris Down
Not true, there are some times where you will have to refer to them with a leading
$ to force the context ($1, $2 ... $N).
shellter
Great, that would certainly affect someones results. I'll update when I get back. Thanks for the improvement!
l0b0
l0b0
You can also use
let myArr[3]++.Increment and update:
array[1]=$((array[1]++))
1 Comment
joharr
Arithmetic expansion obviates the need to reassign the element:
a=(1 2 3); ((a[0]++)); echo ${a[@]}.