Pass length of argument into bash command substitution

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I'm trying to write a quick bash function that populates a README.md with a $1n followed by underscores the length of $1.



The code I found in other stackexchange questions showed that to print a character <n> times, use



printf '=%.0s' 1..<n>


and indeed, this works (obviously replacing <n> with a number).



To create my README.md, I thought the function would look something like this:



make_readme() 
echo "$1
$(printf '=%.0s' 1..$#1)" > README.md


make_readme "Some project"


This, however, produces a file with this text:



Some project
=


As far as I can tell, $#1 within the $(...) is being replaced with the empty string. My guess is that command substitutions get their own argument scopes, and since there are no arguments passed to the substitution, $1 is being replaced with nothing.



I did finally finagle a couple workarounds:



make_readme() 
underline="printf '=%.0s' 1..$#1"
echo "$1
$(eval "$underline")" > README.md



or



make_readme() 
echo "$1" > README.md
printf '=%.0s' 1..$#1 >> README.md



but it seems like there should be a way to do this in one line.









share







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    up vote
    0
    down vote

    favorite












    I'm trying to write a quick bash function that populates a README.md with a $1n followed by underscores the length of $1.



    The code I found in other stackexchange questions showed that to print a character <n> times, use



    printf '=%.0s' 1..<n>


    and indeed, this works (obviously replacing <n> with a number).



    To create my README.md, I thought the function would look something like this:



    make_readme() 
    echo "$1
    $(printf '=%.0s' 1..$#1)" > README.md


    make_readme "Some project"


    This, however, produces a file with this text:



    Some project
    =


    As far as I can tell, $#1 within the $(...) is being replaced with the empty string. My guess is that command substitutions get their own argument scopes, and since there are no arguments passed to the substitution, $1 is being replaced with nothing.



    I did finally finagle a couple workarounds:



    make_readme() 
    underline="printf '=%.0s' 1..$#1"
    echo "$1
    $(eval "$underline")" > README.md



    or



    make_readme() 
    echo "$1" > README.md
    printf '=%.0s' 1..$#1 >> README.md



    but it seems like there should be a way to do this in one line.









    share







    New contributor




    dfoverdx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm trying to write a quick bash function that populates a README.md with a $1n followed by underscores the length of $1.



      The code I found in other stackexchange questions showed that to print a character <n> times, use



      printf '=%.0s' 1..<n>


      and indeed, this works (obviously replacing <n> with a number).



      To create my README.md, I thought the function would look something like this:



      make_readme() 
      echo "$1
      $(printf '=%.0s' 1..$#1)" > README.md


      make_readme "Some project"


      This, however, produces a file with this text:



      Some project
      =


      As far as I can tell, $#1 within the $(...) is being replaced with the empty string. My guess is that command substitutions get their own argument scopes, and since there are no arguments passed to the substitution, $1 is being replaced with nothing.



      I did finally finagle a couple workarounds:



      make_readme() 
      underline="printf '=%.0s' 1..$#1"
      echo "$1
      $(eval "$underline")" > README.md



      or



      make_readme() 
      echo "$1" > README.md
      printf '=%.0s' 1..$#1 >> README.md



      but it seems like there should be a way to do this in one line.









      share







      New contributor




      dfoverdx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I'm trying to write a quick bash function that populates a README.md with a $1n followed by underscores the length of $1.



      The code I found in other stackexchange questions showed that to print a character <n> times, use



      printf '=%.0s' 1..<n>


      and indeed, this works (obviously replacing <n> with a number).



      To create my README.md, I thought the function would look something like this:



      make_readme() 
      echo "$1
      $(printf '=%.0s' 1..$#1)" > README.md


      make_readme "Some project"


      This, however, produces a file with this text:



      Some project
      =


      As far as I can tell, $#1 within the $(...) is being replaced with the empty string. My guess is that command substitutions get their own argument scopes, and since there are no arguments passed to the substitution, $1 is being replaced with nothing.



      I did finally finagle a couple workarounds:



      make_readme() 
      underline="printf '=%.0s' 1..$#1"
      echo "$1
      $(eval "$underline")" > README.md



      or



      make_readme() 
      echo "$1" > README.md
      printf '=%.0s' 1..$#1 >> README.md



      but it seems like there should be a way to do this in one line.







      bash shell-script command-substitution





      share







      New contributor




      dfoverdx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      share







      New contributor




      dfoverdx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      share



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      dfoverdx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 7 mins ago









      dfoverdx

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      New contributor




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      New contributor





      dfoverdx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      dfoverdx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.

























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