问题
If the following example, which sets the IFS environment variable to a line feed character...
IFS=$\'\\n\'
- What does the dollar sign mean exactly?
- What does it do in this specific case?
- Where can I read more on this specific usage (Google doesn\'t allow special characters in searches and I don\'t know what to look for otherwise)?
I know what the IFS environment variable is, and what the \\n character is (line feed), but why not just use the following form:
IFS=\"\\n\" (which does not work)?
For example, if I want to loop through every line of a file and want to use a for loop, I could do this:
for line in (< /path/to/file); do
echo \"Line: $line\"
done
However, this won\'t work right unless IFS is set to a line feed character. To get it to work, I\'d have to do this:
OLDIFS=$IFS
IFS=$\'\\n\'
for line in (< /path/to/file); do
echo \"Line: $line\"
done
IFS=$OLDIFS
Note: I don\'t need another way for doing the same thing, I know many other already... I\'m only curious about that $\'\\n\' and wondered if anyone could give me an explanation on it.
回答1:
Normally bash doesn't interpret escape sequences in string literals. So if you write \n or "\n" or '\n', that's not a linebreak - it's the letter n (in the first case) or a backslash followed by the letter n (in the other two cases).
$'somestring' is a syntax for string literals with escape sequences. So unlike '\n', $'\n' actually is a linebreak.
回答2:
Just to give the construct its official name: strings of the form $'...' are called ANSI C-quoted strings.
That is, as in [ANSI] C strings, backlash escape sequences are recognized and expanded to their literal equivalent (see below for the complete list of supported escape sequences).
After this expansion, $'...' strings behave the same way as '...' strings - i.e., they're treated as literals NOT subject to any [further] shell expansions.
For instance, $'\n' expands to a literal newline character - which is something a regular bash string literal (whether '...' or "...") cannot do.[1]
Another interesting feature is that ANSI C-quoted strings can escape ' (single quotes) as \', which, '...' (regular single-quoted strings) cannot:
echo $'Honey, I\'m home' # OK; this cannot be done with '...'
List of supported escape sequences:
Backslash escape sequences, if present, are decoded as follows:
\a alert (bell)
\b backspace
\e \E an escape character (not ANSI C)
\f form feed
\n newline
\r carriage return
\t horizontal tab
\v vertical tab
\ backslash
\' single quote
\" double quote
\nnn the eight-bit character whose value is the octal value nnn (one to three digits)
\xHH the eight-bit character whose value is the hexadecimal value HH (one or two hex digits)
\uHHHH the Unicode (ISO/IEC 10646) character whose value is the hexadecimal value HHHH (one to four hex digits)
\UHHHHHHHH the Unicode (ISO/IEC 10646) character whose value is the hexadecimal value HHHHHHHH (one to eight hex digits)
\cx a control-x character
The expanded result is single-quoted, as if the dollar sign had not been present.
[1] You can, however, embed actual newlines in '...' and "..." strings; i.e., you can define strings that span multiple lines.
回答3:
From http://www.linuxtopia.org/online_books/bash_guide_for_beginners/sect_03_03.html:
Words in the form "$'STRING'" are treated in a special way. The word expands to a string, with backslash-escaped characters replaced as specified by the ANSI-C standard. Backslash escape sequences can be found in the Bash documentation.found
I guess it's forcing the script to escape the line feed to the proper ANSI-C standard.
回答4:
Re recovering the default IFS- this OLDIFS=$IFS is not necessary. Run new IFS in subshell to avoid overriding the default IFS:
ar=(123 321); ( IFS=$'\n'; echo ${ar[*]} )
Besides I don't really believe you recover the old IFS fully. You should double quote it to avoid line breaking such as OLDIFS="$IFS".
回答5:
ANSI C-quoted strings is a key point. Thanks to @mklement0 .
You can test ANSI C-quoted strings with command od.
echo -n $'\n' | od -c
echo -n '\n' | od -c
echo -n $"\n" | od -c
echo -n "\n" | od -c
Outputs:
0000000 \n
0000001
0000000 \ n
0000002
0000000 \ n
0000002
0000000 \ n
0000002
You can know the meaning clearly by the outputs.
回答6:
It's like retrieving the value from a variable:
VAR='test'
echo VAR
echo $VAR
are different, so the dollar sign basically evaluates the content.
来源:https://stackoverflow.com/questions/4128235/what-is-the-exact-meaning-of-ifs-n