Re: more splitting

[Ray Andrews <rayandrews@xxxxxxxxxxx>]

On 2026-04-14 20:26, Bart Schaefer wrote:

> If you redo 'hex' to accept the name of the variable instead of the
> expansion of the variable, you can fix all of this:
>
> function hex ()
> {
>     local _hex_var element
>     echo
>     if [[ -p /dev/fd/0 ]]; then
>        read -d '' -r _hex_var
>     else
>        _hex_var=$1
>     fi
>     set -- ${(P)_hex_var}
>     print -rC1 -- ${(q+)${(Q)@}}
>     echo "\n-----------------------------\n"
>     for element ("$@") print -rn -- $element | od -vAx -tx1 -tc
> }
>
> hex var
> print var | hex
>     
    Perfect!  If the *name* is passed, then it doesn't matter if it's a
    data stream.  I trust a name is always a string, so front door or
    back door ... all the same.  I hope my comments are correct:
    
      # Test variable:
      var=("a b" c$'\n''d e f'' ''g h' ij)
      
      # % hex var
      # % print var | hex
      # We pass the NAME of the array variable thus the pipe || argument
      methods work exactly the same. If we passed the value of the array
      instead, the pipe would lose all splitting information, turning it
      into a string. Yes? Bart knows these things.
      function hex ()
      {
          local _hex_var element
          echo
          if [[ -p /dev/fd/0 ]]; then  # We are smoking a pipe.
             read -d '' -r _hex_var   # -d:terminate on nul,
      -r:'raw'--all literal.
          else
             _hex_var=$1             # We receive the NAME of the var
      :-) tx. Bart
          fi
          set -- ${(P)_hex_var}        # Expand PPs to value of named
      var.
          # (q+):quote the quotes ... sorta, -C1: columns split on
      spaces.
          print -rC1 -- ${(q+)@}
          echo "\n-----------------------------\n"
          for element ("$@") print -rn -- $element | od -vAx -tx1 -tc
      }
      
      ... but many questions running the thing ...

On Tue, Apr 14, 2026 at 8:26 PM Bart Schaefer <schaefer@xxxxxxxxxxxxxxxx> wrote:

    print -rC1 -- ${(q+)${(Q)@}}

Sorry, there's an extra (Q) in there, should be just
    print -rC1 -- ${(q+)@}

Too much copy-pasting.

That was one of my questions :-)
Already fixed above.