Why IFS did not effect?

Hi,
   I write following scripts:

    IFS=",";set | grep "IFS"; for name in 0.01,0.02; do echo "$name";
done

   And its ouput is   "0.01,0.02", But I think it should be :
  0.01
  0.02

  The IFS variable take no effect. Could you please tell me why?
Thanks!

Regards!
Bo

In article 
< XXXX@XXXXX.COM >,




IFS is only used after variable expansion, not when parsing the original 
command line.  Otherwise, you would have to write:

for,name,in,0.1,0.2

-- 
Barry Margolin,  XXXX@XXXXX.COM 
Arlington, MA
*** PLEASE post questions in newsgroups, not directly to me ***
*** PLEASE don't copy me on replies, I'll read them in the group ***

You need the following:

stuff=0.01,0.02
old_IFS=$IFS
IFS=,
for name in $stuff
do echo "$name"
done
printf 'inner [%s]\n' "$IFS"
IFS=$old_IFS
printf 'after [%s]\n' "$IFS"

notice that,  at least under Bash,  "for" is a  syntax so we
have to  save IFS and  restore it later; the  following will
raise an error "for: command not found":

IFS=, for name in $stuff
do echo "$name"
done

HTH
-- 
Marco Maggi

This is true but makes no sense to me; IFS is used for
word splitting which occurs after expansions.  Token
recognition occurs first and uses blanks to delimit
tokens (it is in reality more complex that this, but
IFS is definitely NOT consulted at this point.)

So with a for loop (compound command), reserved words
should be recognized.  According to the man pages
(and SUS4, Sec. 2.9.4) only the list of words following
the "in" keyword are expanded, and thus subject to IFS,
at least as far as I understand it.

This behavior occurs in ksh too (and still doesn't
work in zsh but that give a different error).

This works:
IFS=, eval 'for i in $stuff; do printf "%s\n"  $i; done'

and you can of course use:
( IFS=,; for i in $stuff; do printf '%s\n' "$i"; done )

But why should you get errors with these:
   $ IFS=, for i in $FOO; do echo "$i"; done;
   -bash: syntax error near unexpected token `do'
and:
   $ IFS=, for i in $FOO
   -bash: for: command not found
(And similarly for ksh.)

What am I missing?

-Wayne

Variable assignment cannot precede a reserved word. What you noted is true
not only for the "for" reserved word, but also for the others:

$ foobar=10 if true; then echo foo; fi
bash: syntax error near unexpected token `then'

$ foobar=10 while true; do echo foo; done
bash: syntax error near unexpected token `do'

etc.

According to posix, reserved words are:

!     do     esac   in
{     done   fi     then
}     elif   for    until
case  else   if     while

and reserved words are recognized as follows:

"This recognition shall only occur when none of the characters is quoted and
when the word is used as:

    * The first word of a command

    * The first word following one of the reserved words other than case,
for, or in
    
    * The third word in a case command (only in is valid in this case)
    
    * The third word in a for command (only in and do are valid in this
case)"

so if a variable assignment is at the beginning of the line, the reserved
word is not recognized.
BTW, also not that they must not be quoted:

$ "for" i in a; do echo $i; done
bash: syntax error near unexpected token `do'


Maybe you want this:

for i in \; ; do echo $i; done


That is just normal syntax if you want to write the for command all on a
single line:

for i in $FOO; do echo $i; done

(module the variable assignment at the beginning which is not valid before a
reserved word)
 

Same as before.