Very Computer

Hi!
I am translating man 1 bash in my native languge. There is a somewhat confu-
sing explanation (at leas for me) of IFS varible in EXPANSION section, Word
Splitting subsection. I'll quote it here:

 >  Word Splitting
 >      The shell scans the results of parameter expansion, command
substitu-
 >      tion,  and  arithmetic  expansion  that  did  not occur within
double
 >      quotes for word splitting."
example:
var="a  b c"
echo $var

 >      The shell treats each character of IFS as a delimiter, and
splits the
 >      results  of  the other expansions into words on these
characters.  If
 >      IFS is unset, or its  value  is  exactly  <space><tab><newline>,
  the
 >      default, then any sequence of IFS characters serves to delimit
words.
this is clear.

 >      If IFS has a value other than the  default,  then  sequences  of
  the
 >      whitespace  characters space and tab are ignored at the
beginning and
 >      end of the word, as long as the whitespace character is in the
value

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
 >      of  IFS  (an IFS whitespace character).
that means, *unless the witespace character is back in the value of IFS*

 >                                               Any character in IFS
that is
 >      not IFS whitespace, along with any adjacent  IFS  whitespace
charac-
 >      ters,  delimits  a field.
I think, I got it:
IFS=":;"
var="a:;b;::c"
echo $var   # a  b   c

Now the part that I can barely understand:
 >                                A sequence of IFS whitespace characters is
 >      also treated as a delimiter.  If the value of IFS is  null,  no
  word
 >      splitting occurs.
 >
 >      Explicit  null  arguments ("" or '') are retained.  Unquoted
implicit
 >      null arguments, resulting from the expansion of parameters that
  have
 >      no  values,  are  removed.   If a parameter with no value is
expanded
 >      within double quotes, a null argument results and is retained.
 >
 >      Note that if no expansion occurs, no splitting is performed.
Could someone, please, expand this last part, add a few examples.

Thank you. Vitaliy.

    Multiple adjacent whitespace characters are considered a single
    delimiter, but multiple non-whitespace characters each delimit a
    field.

    In your example, "a:;b;::c" contains 6 fields, including 3 empty
    fields. But "a:;b c" contains only 3 ("a", "", and "b c") or 4 if
    you add a space to IFS.

    You can see the splitting more clearly if you use printf instead
    of echo:

printf "%s\n" $var

--
    Chris F.A. Johnson                     <http://cfaj.freeshell.org>
    ==================================================================
    Shell Scripting Recipes: A Problem-Solution Approach, 2005, Apress
    <http://www.torfree.net/~chris/books/cfaj/ssr.html>

No, that means that whitespace characters in IFS are treated differently
from printable characters. If a whitespace character occurs in the IFS,
solely, or in combination with other characters (:; in your example)
contiguous whitespace characters are collapsed into a single whitespace
as far as word-splitting is concerned. Sequences of non-whitespace
characters are not collapsed.

examples:

~>a="a  :       b:::c"

~>printf "\"%s\"\n" $a
"a"
":"
"b:::c"

~>IFS=":"

~>printf "\"%s\"\n" $a
"a  "
"       b"
""
""
"c"

~>IFS=":        "

~>printf "\"%s\"\n" $a
"a"
"b"
""
""
"c"

Quote:> Now the part that I can barely understand:
> > A sequence of IFS whitespace characters is also treated as a delimiter.  

A sequence of whitespace characters is collapsed into a single delimiter,
See previous examples.

Quote:> > If the value of IFS is  null, no word splitting occurs.

try it:

~>IFS=""

~>printf "\"%s\"\n" $a
"a  :       b:::c"