bash fill an array of integers from arguments and sort it

Question

I looked everybody but I'm stuck with this code.

For example : The user calls ./array.sh 3 5 6 2 1 I am supposed to sort (i thought bubble sort) and print the array sorted.

    #! /bin/bash
    tab=( $@ )
    define -i temp
    for ((i = ${#tab[*]-1 ; i >= 0 ; i--)) ; do
        for ((j = 0 ; i - 1; j++)) ; do
        if [ ${tab[$i]} > ${tab[$i+1]} ] then
            $temp = ${tab[$i+1]
            ${tab[$i+1]} = ${tab[$i]}
            ${$tab[$i]} = $temp 
        fi
    done
    echo ${tab[*]} #print the array

But bash is not happy with that, he keeps tellin me that I cannot assing values like that. What do I do wrong ? Can you help me please ? I looked in a lot of places but there is no way to find the solution.

Thanks you in advance guys.

Run your script through shellcheck.net; most of your problems are simple syntax errors. — chepner
– chepner, Commented Oct 15, 2014 at 19:29
I strongly urge you to fix your errors: that will help your development as a programmer. I also urge you not to reinvent the wheel: printf "%s\n" "$@" | sort -n suffices. — glenn jackman
– glenn jackman, Commented Oct 15, 2014 at 19:36
chepner thank you very much for the website, I almost corrected the code when @JohnBollinger showed me the way. for GlennJackman , I know right, but that's an assignment from a very old school teacher, thanks again everybody — beorn
– beorn, Commented Oct 15, 2014 at 20:16

John Bollinger · Accepted Answer · 2014-10-15 20:17:39Z

2

You cannot assign a value to a value (e.g. ${tab[$i+1]} = ${tab[$i]}); you must assign to a name or to an array element. Also, you may not have space on either side of the equals sign (=) in a bash assignment, except in numeric context.

And you dropped some closing braces.

And you misspelled "declare".

And you need a semicolon or newline between your if condition and your then.

And you didn't terminate your inner loop.

And you reference a non-existent array element via ${tab[$i+1]}.

And your inner loop has a constant as its termination condition.

And ${#tab[*]-1} incorrectly attempts to do math inside the braces delimiting the variable reference.

And you referenced the wrong index variable (consistently) in your inner loop.

And > is a redirection operator, not greater than, except in numeric context.

Once you clear up that multitude of errors, you end up with

#! /bin/bash
tab=( $@ )
declare -i temp
for ((i = ${#tab[*]} - 1; i > 0 ; i--)) ; do
  for ((j = 0 ; $j < $i; j++)) ; do
    if [ ${tab[$j]} -gt ${tab[$j+1]} ]; then
      temp=${tab[$j+1]}
      tab[$j+1]=${tab[$j]}
      tab[$j]=$temp 
    fi
  done
done
echo ${tab[*]} #print the array

which actually works.

edited Oct 15, 2014 at 20:17

answered Oct 15, 2014 at 19:57

John Bollinger

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1 Comment

beorn Over a year ago

thank you a lot, i am a beginner and i was used to code in c, so bash is really new for me, sorry for wasting your time but I really looked in a lot of places even here. thanks again a lot

Tom Fenech · Accepted Answer · 2014-10-15 19:40:04Z

2

Your variable assignment syntax is all wrong.

First, you don't put $ before the variable being assigned.

Second, you must not have spaces around =.

So it should be:

temp=${tab[$i+1]}
tab[$i+1]=${tab[$i]}
tab[$i]=$temp

edited Oct 15, 2014 at 19:40

Tom Fenech

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answered Oct 15, 2014 at 19:30

Barmar

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Comments

John1024 · Accepted Answer · 2014-10-15 20:18:14Z

1

I understand that you are working on a class assignment or such and are restricted in how you are allowed to solve the problem. For those not so restricted, here is a simple solution using standard unix tools:

#!/bin/bash
( IFS=$'\n'; echo "$*" ) | sort -n

Sample usage:

$ script.sh 3 5 6 2 1
1
2
3
5
6

Explanation:

( IFS=$'\n'; echo "$*" )

This causes the command line arguments to be printed, one per line. This is in a subshell so that the assignment to IFS does not affect the rest of the script.
sort -n

-n tells sort to apply a numeric sort.

answered Oct 15, 2014 at 20:18

John1024

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