read in bash on whitespace-delimited file without empty fields collapsing

Question

I'm trying to read a multi-line tab-separated file in bash. The format is such that empty fields are expected. Unfortunately, the shell is collapsing together field separators which are next to each other, as so:

# IFS=$''
# read one two three <<<$'onethree'
# printf '<%s> ' "$one" "$two" "$three"; printf '
'
<one> <three> <>

...as opposed to the desired output of <one> <> <three>.

Can this be resolved without resorting to a separate language (such as awk)?

Answer 1

Sure

---

IFS=,
echo $'onethree' | tr \11 , | (
  read one two three
  printf '<%s> ' "$one" "$two" "$three"; printf '
'
)

I've rearranged the example just a bit, but only to make it work in any Posix shell.

Update: Yeah, it seems that white space is special, at least if it's in IFS. See the second half of this paragraph from bash(1):

   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.  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).  Any character in IFS that is not IFS white-
   space, along with any adjacent IFS whitespace  characters,  delimits  a
   field.   A  sequence  of IFS whitespace characters is also treated as a
   delimiter.  If the value of IFS is null, no word splitting occurs.