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Shell scripting question



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Community Moderator Election Results
Why I closed the “Why is Kali so hard” questionionCube shell script gives errorShell script wrapper to prevent running command with no arguments?“too many arguments in [ (test) statement”shell calling another shell program fails when run by crontabshell script works fine when executed in terminal, but errors out when ran as concurrent programHow to loop through the files extracted by a find command and pass is as input to another command?getopts Unix inputOperand expected: syntax error (comparing statement)Shell script renameHow to get du -ksh working without a carriage return in shell-scripting?



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








-1















This Linux program is supposed to print 0 to 4. But I'm unable to find error in this code.



#!/bin/sh

a=0

while [ $a -lt 5 ]
do
echo $a
a='expr $a + 1'
done


When I try to run it, it says 'Line 5: [: too many arguments'.










share|improve this question



















  • 1





    Please give your question a more descriptive title.

    – phk
    Oct 17 '16 at 17:55

















-1















This Linux program is supposed to print 0 to 4. But I'm unable to find error in this code.



#!/bin/sh

a=0

while [ $a -lt 5 ]
do
echo $a
a='expr $a + 1'
done


When I try to run it, it says 'Line 5: [: too many arguments'.










share|improve this question



















  • 1





    Please give your question a more descriptive title.

    – phk
    Oct 17 '16 at 17:55













-1












-1








-1








This Linux program is supposed to print 0 to 4. But I'm unable to find error in this code.



#!/bin/sh

a=0

while [ $a -lt 5 ]
do
echo $a
a='expr $a + 1'
done


When I try to run it, it says 'Line 5: [: too many arguments'.










share|improve this question
















This Linux program is supposed to print 0 to 4. But I'm unable to find error in this code.



#!/bin/sh

a=0

while [ $a -lt 5 ]
do
echo $a
a='expr $a + 1'
done


When I try to run it, it says 'Line 5: [: too many arguments'.







shell-script






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 13 at 15:07









Rui F Ribeiro

42.2k1484142




42.2k1484142










asked Oct 17 '16 at 13:46









SparshSparsh

4




4







  • 1





    Please give your question a more descriptive title.

    – phk
    Oct 17 '16 at 17:55












  • 1





    Please give your question a more descriptive title.

    – phk
    Oct 17 '16 at 17:55







1




1





Please give your question a more descriptive title.

– phk
Oct 17 '16 at 17:55





Please give your question a more descriptive title.

– phk
Oct 17 '16 at 17:55










3 Answers
3






active

oldest

votes


















4














You have use to backtick quotes for command substitution



a=`expr $a + 1`





share|improve this answer




















  • 5





    Or better now adays is $(...) notation

    – Eric Renouf
    Oct 17 '16 at 13:54











  • @Eric Renouf Is this portable?

    – rudimeier
    Oct 17 '16 at 13:57






  • 1





    Yes, it's specified as part of the POSIX shell command language: pubs.opengroup.org/onlinepubs/9699919799

    – Eric Renouf
    Oct 17 '16 at 14:02







  • 2





    stackoverflow.com/questions/4708549/… apparently yes, and $(...) is better practice because it makes nesting clearer (whereas backticks need escaping for nesting).

    – Valentin B.
    Oct 17 '16 at 14:03


















2














I've provided three examples, first two are with while loops and the third is with an until loop. The first example uses very close syntax to your original, the third is a bit more complex as it uses dynamic settings for counter maximum and an array for iterating though words (space separated arguments) passed one at a time. Now these are simple examples and intended to only answer your direct questions in the scripting language of Bash if you want more advanced examples see the link attached to a script I'm maintaining on GitHub for examples of nested loops among other tech-wizardry.



Example one



#!/usr/bin/env bash
let _counter=0
let _max="5"
while [ "$_counter" -lt "$_max" ]; do
echo "$_counter"
let _counter=++
done
unset _counter
unset _max


Running example one example



./example_one.sh
1
2
3
4
5


Example two



#!/usr/bin/env bash
_arr_args=( "$@" )
let _counter=0
let _max="$#_arr_args[@]"
while [ "$_counter" -lt "$_max" ]; do
echo "$_arr_args[$_counter]"
let _counter=++
done
unset _counter
unset _max


Running example two example



./example_two.sh "$(seq 1 5)"
## same output as example one


Example three (bonus)



#!/usr/bin/env bash
_arr_args=( "$@" )
let _counter=0
let _max="$#_arr_args[@]"
until [ "$_counter" = "$_max" ]; do
echo "$_arr_args[$_counter]"
let _counter=++
done
unset _counter
unset _max


Running example three example



./example_three.sh "$(seq 1 5)"
## same output as example one


Notes for new Bash script writers



  • look up built-in variables/arrays and substitutions rules; it'll save you time, effort, and from unneeded piping to other programs.

  • look up shellcheck it's available in most distributions, and in source form from it's authors, and is integrated into Code Climate with GitHub.

  • as eluded to above here is the link to an example of a script that makes use of both looping types shown above as well as auto code checking. However, the scripts' intention is well beyond the scope of answering the above question so search for "while" & "until" to find the relevant functions that use these calls.


  • bash != sh there are similarities but they're different, personally I find bash to be the better choice because it's less likely to bugger-up if given a sh script where as shell will do odd things with bash syntax.





share|improve this answer
































    0














    You are assigning the string 'expr $a + 1' to a. That's not likely what you want. Just change single quotes to backticks and it will work:



    a=`expr $a + 1`


    The above should answer your question. Below is some additional thoughts:



    If your /bin/sh implementation is Bash, you could also consider using $(...) instead of backticks. This has several advantages: it avoids typographic confusion between the single quote and the backtick, the $(...) form can be nested, and each nesting is in its own quoting context (so you can do things such as:



    variable="$(somecommand -o "$file")"


    Here is another implementation:



    #/bin/bash

    declare -i a=0
    while [[ $a -lt 5 ]]; do
    echo $a
    ((a++))
    done


    Note that sh is not an implementation. So, to answer your question safely, we'd need to know if your intention includes POSIX portability, or if you are interested in a particular implementation such as Bash, dash, tcsh, etc.






    share|improve this answer

























    • $(...) is POSIX, it will work in any POSIX shell (so the POSIX sh of any system, most of the time found in /bin), not just bash $((...)) arithmetic expansion is also POSIX.

      – Stéphane Chazelas
      Oct 17 '16 at 14:09












    • @ValentinB. $(...) is POSIX but not Bourne. Every system will have a sh where that works but not necessarily in /bin like on Solaris 10 and older where /bin/sh is the Bourne shell (an ancient, non-POSIX shell), and the standard shell is in /usr/xpg4/bin

      – Stéphane Chazelas
      Oct 17 '16 at 14:11












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    You have use to backtick quotes for command substitution



    a=`expr $a + 1`





    share|improve this answer




















    • 5





      Or better now adays is $(...) notation

      – Eric Renouf
      Oct 17 '16 at 13:54











    • @Eric Renouf Is this portable?

      – rudimeier
      Oct 17 '16 at 13:57






    • 1





      Yes, it's specified as part of the POSIX shell command language: pubs.opengroup.org/onlinepubs/9699919799

      – Eric Renouf
      Oct 17 '16 at 14:02







    • 2





      stackoverflow.com/questions/4708549/… apparently yes, and $(...) is better practice because it makes nesting clearer (whereas backticks need escaping for nesting).

      – Valentin B.
      Oct 17 '16 at 14:03















    4














    You have use to backtick quotes for command substitution



    a=`expr $a + 1`





    share|improve this answer




















    • 5





      Or better now adays is $(...) notation

      – Eric Renouf
      Oct 17 '16 at 13:54











    • @Eric Renouf Is this portable?

      – rudimeier
      Oct 17 '16 at 13:57






    • 1





      Yes, it's specified as part of the POSIX shell command language: pubs.opengroup.org/onlinepubs/9699919799

      – Eric Renouf
      Oct 17 '16 at 14:02







    • 2





      stackoverflow.com/questions/4708549/… apparently yes, and $(...) is better practice because it makes nesting clearer (whereas backticks need escaping for nesting).

      – Valentin B.
      Oct 17 '16 at 14:03













    4












    4








    4







    You have use to backtick quotes for command substitution



    a=`expr $a + 1`





    share|improve this answer















    You have use to backtick quotes for command substitution



    a=`expr $a + 1`






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Oct 17 '16 at 14:01

























    answered Oct 17 '16 at 13:51









    rudimeierrudimeier

    5,7321832




    5,7321832







    • 5





      Or better now adays is $(...) notation

      – Eric Renouf
      Oct 17 '16 at 13:54











    • @Eric Renouf Is this portable?

      – rudimeier
      Oct 17 '16 at 13:57






    • 1





      Yes, it's specified as part of the POSIX shell command language: pubs.opengroup.org/onlinepubs/9699919799

      – Eric Renouf
      Oct 17 '16 at 14:02







    • 2





      stackoverflow.com/questions/4708549/… apparently yes, and $(...) is better practice because it makes nesting clearer (whereas backticks need escaping for nesting).

      – Valentin B.
      Oct 17 '16 at 14:03












    • 5





      Or better now adays is $(...) notation

      – Eric Renouf
      Oct 17 '16 at 13:54











    • @Eric Renouf Is this portable?

      – rudimeier
      Oct 17 '16 at 13:57






    • 1





      Yes, it's specified as part of the POSIX shell command language: pubs.opengroup.org/onlinepubs/9699919799

      – Eric Renouf
      Oct 17 '16 at 14:02







    • 2





      stackoverflow.com/questions/4708549/… apparently yes, and $(...) is better practice because it makes nesting clearer (whereas backticks need escaping for nesting).

      – Valentin B.
      Oct 17 '16 at 14:03







    5




    5





    Or better now adays is $(...) notation

    – Eric Renouf
    Oct 17 '16 at 13:54





    Or better now adays is $(...) notation

    – Eric Renouf
    Oct 17 '16 at 13:54













    @Eric Renouf Is this portable?

    – rudimeier
    Oct 17 '16 at 13:57





    @Eric Renouf Is this portable?

    – rudimeier
    Oct 17 '16 at 13:57




    1




    1





    Yes, it's specified as part of the POSIX shell command language: pubs.opengroup.org/onlinepubs/9699919799

    – Eric Renouf
    Oct 17 '16 at 14:02






    Yes, it's specified as part of the POSIX shell command language: pubs.opengroup.org/onlinepubs/9699919799

    – Eric Renouf
    Oct 17 '16 at 14:02





    2




    2





    stackoverflow.com/questions/4708549/… apparently yes, and $(...) is better practice because it makes nesting clearer (whereas backticks need escaping for nesting).

    – Valentin B.
    Oct 17 '16 at 14:03





    stackoverflow.com/questions/4708549/… apparently yes, and $(...) is better practice because it makes nesting clearer (whereas backticks need escaping for nesting).

    – Valentin B.
    Oct 17 '16 at 14:03













    2














    I've provided three examples, first two are with while loops and the third is with an until loop. The first example uses very close syntax to your original, the third is a bit more complex as it uses dynamic settings for counter maximum and an array for iterating though words (space separated arguments) passed one at a time. Now these are simple examples and intended to only answer your direct questions in the scripting language of Bash if you want more advanced examples see the link attached to a script I'm maintaining on GitHub for examples of nested loops among other tech-wizardry.



    Example one



    #!/usr/bin/env bash
    let _counter=0
    let _max="5"
    while [ "$_counter" -lt "$_max" ]; do
    echo "$_counter"
    let _counter=++
    done
    unset _counter
    unset _max


    Running example one example



    ./example_one.sh
    1
    2
    3
    4
    5


    Example two



    #!/usr/bin/env bash
    _arr_args=( "$@" )
    let _counter=0
    let _max="$#_arr_args[@]"
    while [ "$_counter" -lt "$_max" ]; do
    echo "$_arr_args[$_counter]"
    let _counter=++
    done
    unset _counter
    unset _max


    Running example two example



    ./example_two.sh "$(seq 1 5)"
    ## same output as example one


    Example three (bonus)



    #!/usr/bin/env bash
    _arr_args=( "$@" )
    let _counter=0
    let _max="$#_arr_args[@]"
    until [ "$_counter" = "$_max" ]; do
    echo "$_arr_args[$_counter]"
    let _counter=++
    done
    unset _counter
    unset _max


    Running example three example



    ./example_three.sh "$(seq 1 5)"
    ## same output as example one


    Notes for new Bash script writers



    • look up built-in variables/arrays and substitutions rules; it'll save you time, effort, and from unneeded piping to other programs.

    • look up shellcheck it's available in most distributions, and in source form from it's authors, and is integrated into Code Climate with GitHub.

    • as eluded to above here is the link to an example of a script that makes use of both looping types shown above as well as auto code checking. However, the scripts' intention is well beyond the scope of answering the above question so search for "while" & "until" to find the relevant functions that use these calls.


    • bash != sh there are similarities but they're different, personally I find bash to be the better choice because it's less likely to bugger-up if given a sh script where as shell will do odd things with bash syntax.





    share|improve this answer





























      2














      I've provided three examples, first two are with while loops and the third is with an until loop. The first example uses very close syntax to your original, the third is a bit more complex as it uses dynamic settings for counter maximum and an array for iterating though words (space separated arguments) passed one at a time. Now these are simple examples and intended to only answer your direct questions in the scripting language of Bash if you want more advanced examples see the link attached to a script I'm maintaining on GitHub for examples of nested loops among other tech-wizardry.



      Example one



      #!/usr/bin/env bash
      let _counter=0
      let _max="5"
      while [ "$_counter" -lt "$_max" ]; do
      echo "$_counter"
      let _counter=++
      done
      unset _counter
      unset _max


      Running example one example



      ./example_one.sh
      1
      2
      3
      4
      5


      Example two



      #!/usr/bin/env bash
      _arr_args=( "$@" )
      let _counter=0
      let _max="$#_arr_args[@]"
      while [ "$_counter" -lt "$_max" ]; do
      echo "$_arr_args[$_counter]"
      let _counter=++
      done
      unset _counter
      unset _max


      Running example two example



      ./example_two.sh "$(seq 1 5)"
      ## same output as example one


      Example three (bonus)



      #!/usr/bin/env bash
      _arr_args=( "$@" )
      let _counter=0
      let _max="$#_arr_args[@]"
      until [ "$_counter" = "$_max" ]; do
      echo "$_arr_args[$_counter]"
      let _counter=++
      done
      unset _counter
      unset _max


      Running example three example



      ./example_three.sh "$(seq 1 5)"
      ## same output as example one


      Notes for new Bash script writers



      • look up built-in variables/arrays and substitutions rules; it'll save you time, effort, and from unneeded piping to other programs.

      • look up shellcheck it's available in most distributions, and in source form from it's authors, and is integrated into Code Climate with GitHub.

      • as eluded to above here is the link to an example of a script that makes use of both looping types shown above as well as auto code checking. However, the scripts' intention is well beyond the scope of answering the above question so search for "while" & "until" to find the relevant functions that use these calls.


      • bash != sh there are similarities but they're different, personally I find bash to be the better choice because it's less likely to bugger-up if given a sh script where as shell will do odd things with bash syntax.





      share|improve this answer



























        2












        2








        2







        I've provided three examples, first two are with while loops and the third is with an until loop. The first example uses very close syntax to your original, the third is a bit more complex as it uses dynamic settings for counter maximum and an array for iterating though words (space separated arguments) passed one at a time. Now these are simple examples and intended to only answer your direct questions in the scripting language of Bash if you want more advanced examples see the link attached to a script I'm maintaining on GitHub for examples of nested loops among other tech-wizardry.



        Example one



        #!/usr/bin/env bash
        let _counter=0
        let _max="5"
        while [ "$_counter" -lt "$_max" ]; do
        echo "$_counter"
        let _counter=++
        done
        unset _counter
        unset _max


        Running example one example



        ./example_one.sh
        1
        2
        3
        4
        5


        Example two



        #!/usr/bin/env bash
        _arr_args=( "$@" )
        let _counter=0
        let _max="$#_arr_args[@]"
        while [ "$_counter" -lt "$_max" ]; do
        echo "$_arr_args[$_counter]"
        let _counter=++
        done
        unset _counter
        unset _max


        Running example two example



        ./example_two.sh "$(seq 1 5)"
        ## same output as example one


        Example three (bonus)



        #!/usr/bin/env bash
        _arr_args=( "$@" )
        let _counter=0
        let _max="$#_arr_args[@]"
        until [ "$_counter" = "$_max" ]; do
        echo "$_arr_args[$_counter]"
        let _counter=++
        done
        unset _counter
        unset _max


        Running example three example



        ./example_three.sh "$(seq 1 5)"
        ## same output as example one


        Notes for new Bash script writers



        • look up built-in variables/arrays and substitutions rules; it'll save you time, effort, and from unneeded piping to other programs.

        • look up shellcheck it's available in most distributions, and in source form from it's authors, and is integrated into Code Climate with GitHub.

        • as eluded to above here is the link to an example of a script that makes use of both looping types shown above as well as auto code checking. However, the scripts' intention is well beyond the scope of answering the above question so search for "while" & "until" to find the relevant functions that use these calls.


        • bash != sh there are similarities but they're different, personally I find bash to be the better choice because it's less likely to bugger-up if given a sh script where as shell will do odd things with bash syntax.





        share|improve this answer















        I've provided three examples, first two are with while loops and the third is with an until loop. The first example uses very close syntax to your original, the third is a bit more complex as it uses dynamic settings for counter maximum and an array for iterating though words (space separated arguments) passed one at a time. Now these are simple examples and intended to only answer your direct questions in the scripting language of Bash if you want more advanced examples see the link attached to a script I'm maintaining on GitHub for examples of nested loops among other tech-wizardry.



        Example one



        #!/usr/bin/env bash
        let _counter=0
        let _max="5"
        while [ "$_counter" -lt "$_max" ]; do
        echo "$_counter"
        let _counter=++
        done
        unset _counter
        unset _max


        Running example one example



        ./example_one.sh
        1
        2
        3
        4
        5


        Example two



        #!/usr/bin/env bash
        _arr_args=( "$@" )
        let _counter=0
        let _max="$#_arr_args[@]"
        while [ "$_counter" -lt "$_max" ]; do
        echo "$_arr_args[$_counter]"
        let _counter=++
        done
        unset _counter
        unset _max


        Running example two example



        ./example_two.sh "$(seq 1 5)"
        ## same output as example one


        Example three (bonus)



        #!/usr/bin/env bash
        _arr_args=( "$@" )
        let _counter=0
        let _max="$#_arr_args[@]"
        until [ "$_counter" = "$_max" ]; do
        echo "$_arr_args[$_counter]"
        let _counter=++
        done
        unset _counter
        unset _max


        Running example three example



        ./example_three.sh "$(seq 1 5)"
        ## same output as example one


        Notes for new Bash script writers



        • look up built-in variables/arrays and substitutions rules; it'll save you time, effort, and from unneeded piping to other programs.

        • look up shellcheck it's available in most distributions, and in source form from it's authors, and is integrated into Code Climate with GitHub.

        • as eluded to above here is the link to an example of a script that makes use of both looping types shown above as well as auto code checking. However, the scripts' intention is well beyond the scope of answering the above question so search for "while" & "until" to find the relevant functions that use these calls.


        • bash != sh there are similarities but they're different, personally I find bash to be the better choice because it's less likely to bugger-up if given a sh script where as shell will do odd things with bash syntax.






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Apr 15 at 1:39

























        answered Oct 17 '16 at 14:43









        S0AndS0S0AndS0

        1967




        1967





















            0














            You are assigning the string 'expr $a + 1' to a. That's not likely what you want. Just change single quotes to backticks and it will work:



            a=`expr $a + 1`


            The above should answer your question. Below is some additional thoughts:



            If your /bin/sh implementation is Bash, you could also consider using $(...) instead of backticks. This has several advantages: it avoids typographic confusion between the single quote and the backtick, the $(...) form can be nested, and each nesting is in its own quoting context (so you can do things such as:



            variable="$(somecommand -o "$file")"


            Here is another implementation:



            #/bin/bash

            declare -i a=0
            while [[ $a -lt 5 ]]; do
            echo $a
            ((a++))
            done


            Note that sh is not an implementation. So, to answer your question safely, we'd need to know if your intention includes POSIX portability, or if you are interested in a particular implementation such as Bash, dash, tcsh, etc.






            share|improve this answer

























            • $(...) is POSIX, it will work in any POSIX shell (so the POSIX sh of any system, most of the time found in /bin), not just bash $((...)) arithmetic expansion is also POSIX.

              – Stéphane Chazelas
              Oct 17 '16 at 14:09












            • @ValentinB. $(...) is POSIX but not Bourne. Every system will have a sh where that works but not necessarily in /bin like on Solaris 10 and older where /bin/sh is the Bourne shell (an ancient, non-POSIX shell), and the standard shell is in /usr/xpg4/bin

              – Stéphane Chazelas
              Oct 17 '16 at 14:11
















            0














            You are assigning the string 'expr $a + 1' to a. That's not likely what you want. Just change single quotes to backticks and it will work:



            a=`expr $a + 1`


            The above should answer your question. Below is some additional thoughts:



            If your /bin/sh implementation is Bash, you could also consider using $(...) instead of backticks. This has several advantages: it avoids typographic confusion between the single quote and the backtick, the $(...) form can be nested, and each nesting is in its own quoting context (so you can do things such as:



            variable="$(somecommand -o "$file")"


            Here is another implementation:



            #/bin/bash

            declare -i a=0
            while [[ $a -lt 5 ]]; do
            echo $a
            ((a++))
            done


            Note that sh is not an implementation. So, to answer your question safely, we'd need to know if your intention includes POSIX portability, or if you are interested in a particular implementation such as Bash, dash, tcsh, etc.






            share|improve this answer

























            • $(...) is POSIX, it will work in any POSIX shell (so the POSIX sh of any system, most of the time found in /bin), not just bash $((...)) arithmetic expansion is also POSIX.

              – Stéphane Chazelas
              Oct 17 '16 at 14:09












            • @ValentinB. $(...) is POSIX but not Bourne. Every system will have a sh where that works but not necessarily in /bin like on Solaris 10 and older where /bin/sh is the Bourne shell (an ancient, non-POSIX shell), and the standard shell is in /usr/xpg4/bin

              – Stéphane Chazelas
              Oct 17 '16 at 14:11














            0












            0








            0







            You are assigning the string 'expr $a + 1' to a. That's not likely what you want. Just change single quotes to backticks and it will work:



            a=`expr $a + 1`


            The above should answer your question. Below is some additional thoughts:



            If your /bin/sh implementation is Bash, you could also consider using $(...) instead of backticks. This has several advantages: it avoids typographic confusion between the single quote and the backtick, the $(...) form can be nested, and each nesting is in its own quoting context (so you can do things such as:



            variable="$(somecommand -o "$file")"


            Here is another implementation:



            #/bin/bash

            declare -i a=0
            while [[ $a -lt 5 ]]; do
            echo $a
            ((a++))
            done


            Note that sh is not an implementation. So, to answer your question safely, we'd need to know if your intention includes POSIX portability, or if you are interested in a particular implementation such as Bash, dash, tcsh, etc.






            share|improve this answer















            You are assigning the string 'expr $a + 1' to a. That's not likely what you want. Just change single quotes to backticks and it will work:



            a=`expr $a + 1`


            The above should answer your question. Below is some additional thoughts:



            If your /bin/sh implementation is Bash, you could also consider using $(...) instead of backticks. This has several advantages: it avoids typographic confusion between the single quote and the backtick, the $(...) form can be nested, and each nesting is in its own quoting context (so you can do things such as:



            variable="$(somecommand -o "$file")"


            Here is another implementation:



            #/bin/bash

            declare -i a=0
            while [[ $a -lt 5 ]]; do
            echo $a
            ((a++))
            done


            Note that sh is not an implementation. So, to answer your question safely, we'd need to know if your intention includes POSIX portability, or if you are interested in a particular implementation such as Bash, dash, tcsh, etc.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Oct 17 '16 at 14:10

























            answered Oct 17 '16 at 14:03









            Steve AmerigeSteve Amerige

            1518




            1518












            • $(...) is POSIX, it will work in any POSIX shell (so the POSIX sh of any system, most of the time found in /bin), not just bash $((...)) arithmetic expansion is also POSIX.

              – Stéphane Chazelas
              Oct 17 '16 at 14:09












            • @ValentinB. $(...) is POSIX but not Bourne. Every system will have a sh where that works but not necessarily in /bin like on Solaris 10 and older where /bin/sh is the Bourne shell (an ancient, non-POSIX shell), and the standard shell is in /usr/xpg4/bin

              – Stéphane Chazelas
              Oct 17 '16 at 14:11


















            • $(...) is POSIX, it will work in any POSIX shell (so the POSIX sh of any system, most of the time found in /bin), not just bash $((...)) arithmetic expansion is also POSIX.

              – Stéphane Chazelas
              Oct 17 '16 at 14:09












            • @ValentinB. $(...) is POSIX but not Bourne. Every system will have a sh where that works but not necessarily in /bin like on Solaris 10 and older where /bin/sh is the Bourne shell (an ancient, non-POSIX shell), and the standard shell is in /usr/xpg4/bin

              – Stéphane Chazelas
              Oct 17 '16 at 14:11

















            $(...) is POSIX, it will work in any POSIX shell (so the POSIX sh of any system, most of the time found in /bin), not just bash $((...)) arithmetic expansion is also POSIX.

            – Stéphane Chazelas
            Oct 17 '16 at 14:09






            $(...) is POSIX, it will work in any POSIX shell (so the POSIX sh of any system, most of the time found in /bin), not just bash $((...)) arithmetic expansion is also POSIX.

            – Stéphane Chazelas
            Oct 17 '16 at 14:09














            @ValentinB. $(...) is POSIX but not Bourne. Every system will have a sh where that works but not necessarily in /bin like on Solaris 10 and older where /bin/sh is the Bourne shell (an ancient, non-POSIX shell), and the standard shell is in /usr/xpg4/bin

            – Stéphane Chazelas
            Oct 17 '16 at 14:11






            @ValentinB. $(...) is POSIX but not Bourne. Every system will have a sh where that works but not necessarily in /bin like on Solaris 10 and older where /bin/sh is the Bourne shell (an ancient, non-POSIX shell), and the standard shell is in /usr/xpg4/bin

            – Stéphane Chazelas
            Oct 17 '16 at 14:11


















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