How to increment a variable in bash?

Question

I have this script in jQuery, it scrolls to wanted section but with spacing to the top. It works fine on all tested browsers ... except firefox.

$('nav > *').click(function(event){
    event.preventDefault();
    var navClicked = $(this).index();
    var elem = $(this).attr("href");
    $('body').scrollTop($(elem).offset().top - 48);
});

https://jsfiddle.net/5L3xyuuv/1/

Solution 1

There is more than one way to increment a variable in bash, but what you tried will not work.

You can use, for example, arithmetic expansion:

var=$((var+1))
((var=var+1))
((var+=1))
((var++))

Or you can use let:

let "var=var+1"
let "var+=1"
let "var++"

See also: https://tldp.org/LDP/abs/html/dblparens.html.

Solution 2

var=$((var + 1))

Arithmetic in bash uses $((...)) syntax.

Solution 3

Various options to increment by 1, and performance analysis

Thanks to Radu Rădeanu's answer that provides the following ways to increment a variable in bash:

var=$((var+1))
((var=var+1))
((var+=1))
((var++))
let "var=var+1"
let "var+=1" 
let "var++"

There are other ways too. For example, look in the other answers on this question.

let var++
var=$((var++))
((++var))
{
    declare -i var
    var=var+1
    var+=1
}
{
    i=0
    i=$(expr $i + 1)
}

Having so many options leads to these two questions:

1. Is there a performance difference between them?
2. If so which, which performs best?

Incremental performance test code:

#!/bin/bash
# To focus exclusively on the performance of each type of increment
# statement, we should exclude bash performing while loops from the
# performance measure. So, let's time individual scripts that
# increment $i in their own unique way.
# Declare i as an integer for tests 12 and 13.
echo > t12 'declare -i i; i=i+1'
echo > t13 'declare -i i; i+=1'
# Set i for test 14.
echo > t14 'i=0; i=$(expr $i + 1)'
x=100000
while ((x--)); do
    echo >> t0 'i=$((i+1))'
    echo >> t1 'i=$((i++))'
    echo >> t2 '((i=i+1))'
    echo >> t3 '((i+=1))'
    echo >> t4 '((i++))'
    echo >> t5 '((++i))'
    echo >> t6 'let "i=i+1"'
    echo >> t7 'let "i+=1"'
    echo >> t8 'let "i++"'
    echo >> t9 'let i=i+1'
    echo >> t10 'let i+=1'
    echo >> t11 'let i++'
    echo >> t12 'i=i+1'
    echo >> t13 'i+=1'
    echo >> t14 'i=$(expr $i + 1)'
done
for script in t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10 t11 t12 t13 t14; do
    line1="$(head -1 "$script")"
    printf "%-24s" "$line1"
    { time bash "$script"; } |& grep user
    # Since stderr is being piped to grep above, this will confirm
    # there are no errors from running the command:
    eval "$line1"
    rm "$script"
done

Results:

i=$((i+1))              user    0m0.992s
i=$((i++))              user    0m0.964s
((i=i+1))               user    0m0.760s
((i+=1))                user    0m0.700s
((i++))                 user    0m0.644s
((++i))                 user    0m0.556s
let "i=i+1"             user    0m1.116s
let "i+=1"              user    0m1.100s
let "i++"               user    0m1.008s
let i=i+1               user    0m0.952s
let i+=1                user    0m1.040s
let i++                 user    0m0.820s
declare -i i; i=i+1     user    0m0.528s
declare -i i; i+=1      user    0m0.492s
i=0; i=$(expr $i + 1)   user    0m5.464s

Conclusion:

It seems bash is fastest at performing i+=1 when $i is declared as an integer. let statements seem particularly slow, and expr is by far the slowest because it is not a built into bash.

Solution 4

There's also this:

var=`expr $var + 1`

Take careful note of the spaces and also ` is not '

While Radu's answers, and the comments, are exhaustive and very helpful, they are bash-specific. I know you did specifically ask about bash, but I thought I'd pipe in since I found this question when I was looking to do the same thing using sh in busybox under uCLinux. This portable beyond bash.

Solution 5

If you declare $var as an integer, then what you tried the first time will actually work:

$ declare -i var=5
$ echo $var
5
$ var=$var+1
$ echo $var
6

Reference: Types of variables, Bash Guide for Beginners

View more solutions

Comments

[frnt]

add your html codes too or create a jsfiddle.

[Patrick Hellebrand]

@frnt jsfiddle.net/5L3xyuuv/1

[vsync]

what kind of a test page is that? it's almost empty and doesn't reproduce the problem

[vsync]

why did you do all this javascript anyway if it simply "jumps" to the id location on the page like a regular hash link does?

[Patrick Hellebrand]

because your 'jump' ignores the header height if theres one

[gniourf_gniourf]

or ((++var)) or ((var=var+1)) or ((var+=1)).

[Javier López]

or var=$(expr $var + 1)

[phunehehe]

Curiously, var=0; ((var++)) returns an error code while var=0; ((var++)); ((var++)) does not. Any idea why?

[phunehehe]

@RaduRădeanu I'm sure, the same thing happens in zsh too. Actually maybe I didn't make myself clear. This prints 1 (an error code) instead of 0 (successful): var=0; ((var++)); echo $?.

[Radu Rădeanu]

@phunehehe Look at help '(('. The last line says: Returns 1 if EXPRESSION evaluates to 0; returns 0 otherwise.

[DreadPirateShawn]

I suspect the evaluation of zero as 1 is why @gniourf_gniourf's tip includes ((++var)) but not ((var++)).

[rubo77]

So which is the most compatible way?

[ArtOfWarfare]

Vastly better than the accepted answer. In just 10% as much space, you managed to provide enough examples (one is plenty - nine is overkill to the point when you're just showing off), and you provided us with enough info to know that ((...)) is the key to using arithmetic in bash. I didn't realize that just looking at the accepted answer - I thought there was a weird set of rules about order of operations or something leading to all of the parenthesis in the accepted answer.

[Patrick Hellebrand]

your snippet works fine, but if ill copy it to my snippet, it stops working

[Patrick Hellebrand]

Could it be, that other jquery scripts cause this, but only on firefox?

[frnt]

Then you need to go through your codes, as per your query this works perfectly fine on all browser.

[MatthewRock]

Apparently speed correlates with command length. I wonder whether the commands call the same functions.

[wjandrea]

is it safe to use let var++, without the quotes?

[wjandrea]

You can also use i=$((i+1))

[Radon Rosborough]

If process substitution $(...) is available on this shell, I'd recommend using it instead.

[Daniel Eisenreich]

((val++)) doesn't worked in my case. But if I use ((++val)) this worked. Some suggestions why?

[arielf]

@DanielEisenreich both forms seem to work fine in bash version 4.3.48. Maybe you have an old version? $ val=0; echo $val; ((val++)); echo $val gives me: 0 1 `

[Laurence Renshaw]

let only needs quotes if you put spaces in your expression

[x-yuri]

@arielf You don't observe the behavior since most likely you don't set -e.

[Sandeep]

I'm surprised that $((var+1)) works. I would have expected $(($var+1)).

[x-yuri]

To make (( var++ )) work (not fail when var == 0) with set -e, do : $(( var++ )).

[raygozag]

Not what the OP asked.

[Keith Reynolds]

@raygozag The op asked how to increment. My answer first list many methods. How does that not directly answer the op question? Then it provide time information for each method. All that is, is just extra information to assist the op in choosing which method he wants to use.

[raygozag]

It was already answered, the OP did not ask for performance evaluations, no matter how helpful you want to be, they didn't ask for it.

[Jon Spencer]

I found that on Bash 3.2.57, using declare -i x and then doing x+=1 in a loop (doing other thing), x did not increment more than one time. The let and (()) methods worked fine.

[sastorsl]

@JonSpencer you might be hitting this? unix.stackexchange.com/questions/402750/…

[Timo]

Use a mix of let and arithmetic expanison and you are good to go.

[Sephethus]

Why do absolutely none of these options anywhere work for me? I'm getting things like i++: not found or ++i not found or let not found or any number of errors. WTF?

[Satoshi Nakamoto]

Just to rememeber: ++var is prefix and var++ is postfix, there is a difference.