This issue came up on the Libera #bash IRC channel today:
Between bash 4.4 and 5.0, the definition of "IFS whitespace" has apparently
been expanded:
hobbit:~$ bash-4.4
hobbit:~$ ( s=$'a\r\r\rb'; IFS=$'\r' read -ra a <<< "$s"; declare -p a )
declare -a a=([0]="a" [1]="" [2]="" [3]="b")
hobbit:~$ exit
hobbit:~$ bash-5.0
hobbit:~$ ( s=$'a\r\r\rb'; IFS=$'\r' read -ra a <<< "$s"; declare -p a )
declare -a a=([0]="a" [1]="b")
hobbit:~$ exit
In bash 4.4 and earlier, IFS whitespace is always space + tab + newline.
But in 5.0 and later, it's "whatever the locale's isspace() allows",
along with some kind of 0x00 to 0x7f range check (thanks emanuele6).
Now, that's not necessarily bad, but the man page still says:
... If IFS has a value other than the default, then sequences of
the whitespace characters space, tab, and newline 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 whitespace, along with any
adjacent IFS whitespace characters, delimits a field. A sequence
of IFS whitespace characters is also treated as a delimiter. ...
"IFS whitespace" is not formally defined here, but it *looks* like
the text is saying that IFS whitespace is only space, tab, and newline
(which used to be true).
The man page should clarify this.
