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Comment by sillysaurusx | Hacker News Reader

I was today years old when I learned that bash can do simple arithmetic with $(($i + 1)). Thanks!

If you just want to add/subtract 1, in bash you can do just '((i++))', i.e.:

  [mmh@x670]$ i=0
  [mmh@x670]$ ((i++)); echo $i
  1
  [mmh@x670]$ ((i++)); echo $i
  2

Beware that the above is very -bashy- and purists will rip your head off for using it.

There is a gotcha here if you use:

  set -o errexit

Because *any bash arithmetic expression that evaluates to zero returns the exit code of 1*.

So this works:

  #!/usr/bin/env bash
  set -o errexit
  i=0
  ((++i))
  echo $i # 1 will be printed to stdout

But this doesn't:

  #!/usr/bin/env bash
  set -o errexit
  i=0
  ((i++)) # terminates here with $? of 1
  echo $i

The relevant doc is for `let`:

       let arg [arg ...]
              Each arg is an arithmetic expression to be evaluated (see
              ARITHMETIC EVALUATION above).  If the last arg evaluates
              to 0, let returns 1; 0 is returned otherwise.

https://www.gnu.org/software/bash/manual/bash.html#index-let

Yeah it's best to just use POSIX shell arithmetic, and not be clever, and not the other 2 ways bash has to do it! THat is:

    i=$((i + 1))

The (( )) construct, without the $, is different, and it's also not POSIX. It has the gotchas around the exit code, which a plain assignment doesn't have.

Also, the wooledge wiki is an amazing resource for Bash/shell scripting. To learn more magic see: https://mywiki.wooledge.org/ArithmeticExpression

Whoa. No dollar signs at all?

I’m at Lowes — does ((i+1)) work? Or is it just for incrementing / decrementing (which is still very useful!)

EDIT: thinking this over, does this work for variable substitution too? E.g. ((i)) being equivalent to $i. Not that you’d necessarily want to…

`(( ))` is an arithmetic evaluation block. Its content has to be an arithmetic expression. Arithmetic expressions don't require `$` before simple variable names and some more complex expressions like array indexing.

`(( i + 1 ))` will evaluate the result of adding one to `$i` and then throw away the result. It doesn't do anything useful (other than having a different exit code depending on whether the expression evaluated to 0 or not).

`$(( ... ))` evaluates the expression and then returns its value. ie `i=$(( i + 1 ))` will increment `$i`, just like `(( i++ ))` did.

To add arbitrary numbers, you'd have to do an assignment, like this:

  [mmh@x670]$ ((i=i+2)); echo $i
  6
  [mmh@x670]$ ((i=i+2)); echo $i
  8

It does also matter where the plus signs are:

  $ i=0
  $ echo $((++i))
  1
  $ echo $((i++))
  1
  $ echo $((++i))
  3
  $

((++i)) does add 1 before the command is run, and ((i++)) does add 1 after the command is run.

You can also use the builtin command `let`:

    let count=0
    until "$@"; do
        let wait=2**count
        sleep $wait
        let count=count+1
    done

Or with two brackets instead of four parens:

$[i+4]

And, btw, res=$(command) instead of those pesky backticks res=`command` ,-)

Or, more simply, $((i + 1)).

And it's not only bash; it's feature of POSIX shell.

Or as they did in ancient times $[i+1]

> The old format $[expression] is deprecated and will be removed in upcoming versions of bash.

Don't forget about:

   declare -i i
   i+=1

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