bash read 命令使用空格作为分隔符将行分割成单词，即使空格不在 IFS 中

Question

我正在使用 bash。

$ bash --version
GNU bash, version 4.3.11(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2013 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.
From `help read`

这是 help read 的前几行内容:

$ help read | head
read: read [-ers] [-a array] [-d delim] [-i text] [-n nchars] [-N nchars] [-p prompt] [-t timeout] [-u fd] [name ...]
    Read a line from the standard input and split it into fields.

    Reads a single line from the standard input, or from file descriptor FD
    if the -u option is supplied.  The line is split into fields as with word
    splitting, and the first word is assigned to the first NAME, the second
    word to the second NAME, and so on, with any leftover words assigned to
    the last NAME.  Only the characters found in $IFS are recognized as word
    delimiters.

我的 IFS 只是一个换行符，即 \n。

$ echo $IFS

$ echo $IFS | od -tcx1                                                                                     
0000000  \n
         0a
0000001

由于 IFS 中没有空格，因此我不希望 read 将字符串 "foo bar baz" 拆分为三个单词。但 read 确实将其分成了三个单词。

$ read a b c <<< "foo bar baz"; echo $a; echo $b; echo $c
foo
bar
baz

为什么当我的 IFS 中没有空格时，它会使用空格作为分隔符来分割字符串？

Answer 1

这很可能只是您echo IFS 方式的问题

> IFS=$'\t\n '
> echo $IFS | od -tcx1
0000000  \n
         0a
0000001
> echo -n "$IFS" | od -tcx1
0000000  \t  \n
         09  0a  20
0000003