20

What is the defference between using

echo -e "Hello\nWorld" 

and

echo $"Hello\nWorld" 

don't they both output:

Hello
World
4
  • 1
    Did you try it? The second one outputs Hello\nWorld. You need the -e switch to interpret escape characters.
    – a06e
    Oct 16, 2014 at 21:47
  • 4
    For the second one, I suspect that you meant to write it with single quotes, echo $'Hello\nWorld', which will print out on two lines under bash.
    – John1024
    Oct 16, 2014 at 21:50
  • @becko, yes I tried it and they both output the same exact thing.
    – Mark
    Oct 18, 2014 at 12:22
  • Also relevant: unix.stackexchange.com/questions/189787/… May 24, 2022 at 10:09

1 Answer 1

35

echo -e and echo $'...' are both similar in that they support the following escape sequences:

  \a     alert (bell)
  \b     backspace
  \e
  \E     an escape character
  \f     form feed
  \n     new line
  \r     carriage return
  \t     horizontal tab
  \v     vertical tab
  \\     backslash
  \0nnn  the eight-bit character whose value is the octal value nnn (zero to three octal 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)

They do have differences. In addition to the above, echo -e supports:

  \c     suppress further output
  \0nnn  the eight-bit character whose value is the octal value nnn (zero to three octal digits)

By contrast, $'....' supports:

 \'     single quote
 \"     double quote
 \nnn   the eight-bit character whose value is the octal value nnn (one to three digits)
 \cx    a control-x character

Observe that, between the two, the \c extensions are incompatible:

$ echo -e  'start\n\cIstop'
start
$ echo  $'start\n\cIstop'
start
        stop

For echo -e above, \c suppresses further output, thereby ignoring the Istop. By contrast, for $'...', the \cI is interpreted as a tab.

The visually-similar form: $"..."

By contrast with $'...', the function of $"..." is quite different. It will cause the string it contains to be translated according to the current locale.

The echo -e controversy

echo -e is not universally supported by shells and many regard the -e option as a design mistake. Observe:

$ ls
-e  -n
$ echo *
$ printf "%s\n" *
-e
-n

As you can see, if what you are printing with echo starts with a dash, the results can be unexpected. Unless you are sure that the first string that you will print with echo does not start with a dash, you are likely better off using printf.

For these reasons, the POSIX standard concludes:

New applications are encouraged to use printf instead of echo.

Chet Ramey, who has maintained bash for the last 22 years, agrees:

[N]ew code should use printf.

4
  • 1
    Also worth noting that both Chet and POSIX recommend using printf rather than echo
    – geirha
    Oct 17, 2014 at 7:26
  • @geirha Very good points. Answer updated.
    – John1024
    Oct 17, 2014 at 7:34
  • 2
    The problem of unexpected options from globs (like *) isn't actually specific to echo. It happens with any command that recognizes short or long options (written in the traditional way) and takes filename operands, including ls. Arguably the situation is a little worse with echo since (as your source mentions) echo doesn't typically accept -- to indicate the end of options, whereas one can write commands like ls -- *l. (And really one ought usually to write commands that way, when using globs that begin with * or -, especially in scripts.) Oct 17, 2014 at 8:55
  • Thank you for the thorough answer, I have an excercise for my Operating Systems course and I needed to print out some text using the \n escape sequence, so in fact it doesn't matter if I use echo -e '...', echo $'...', or echo $"..."
    – Mark
    Oct 18, 2014 at 12:27

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