Difference in behaviour between whitespace in IFS vs. non-whitespace in IFS

Tag: shell,posix

I know that read command splits the input by the characters in the IFS. So if I set IFS as a space then input to the read command is split using space as the delimiter. This is shown below.

[[email protected] ~]$ IFS=" "
[[email protected] ~]$ read a b c
 foo bar baz qux
[[email protected] ~]$ echo "$a"
foo
[[email protected] ~]$ echo "$b"
bar
[[email protected] ~]$ echo "$c"
baz qux

I was expecting that if I make change the IFS to a non-whitespace character, say, a colon and use colon as delimiter in my input, there should be no change in behaviour. But this turned out to be wrong. For example, in the output below, echo "$a" is blank. Why?

[[email protected] ~]$ IFS=:
[[email protected] ~]$ read a b c
:foo:bar:baz:qux
[[email protected] ~]$ echo "$a"

[[email protected]tos ~]$ echo "$b"
foo
[[email protected] ~]$ echo "$c"
bar:baz:qux
[[email protected] ~]# echo $c
bar baz qux

And why does the output of echo $c does not contain the colons?

Best How To :

Question 1: In the output below, echo "$a" is blank. Why?

Because whitespace is special as defined in POSIX (2.6.5 Field Splitting):

3a. IFS white space shall be ignored at the beginning and end of the input.

Question 2: And why does the output of echo $c does not contain the colons?

Because bash performs word splitting in accordance with IFS (which is still set to :) before passing bar, baz and qux as separate arguments to echo.

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