echo "-e" doesn't print anything

Question

I'm using GNU bash, version 3.00.15(1)-release (x86_64-redhat-linux-gnu). And this command:

echo "-e"

doesn't print anything. I guess this is because "-e" is one of a valid options of echo command because echo "-n" and echo "-E" (the other two options) also produce empty strings.

The question is how to escape the sequence "-e" for echo to get the natural output ("-e").

score 15 · Answer 1 · answered Apr 06 '10 at 04:06

15

The one true way to print any arbitrary string:

printf "%s" "$vars"

answered Apr 06 '10 at 04:06

nanaya

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Wouldn't it be nice? But this won't work for embedded ANSI Escape sequences, and depends on who the value was assigned. Try `v="$(tput sgr0)"; printf "%s" "$v"` or `v=$(printf '\033[4m'); printf "%s" "$v"`etc. – Andreas Spindler Feb 14 '13 at 10:16
@AndreasSpindler, if you want `%b` rather than `%s`... well, it's available in all POSIX printf implementations, and that gives you the same escape-sequence handling as found in `echo`. – Charles Duffy Nov 16 '14 at 16:50

score 13 · Accepted Answer · 2008-12-25T21:13:06.177

13

This is a tough one ;)

Usually you would use double dashes to tell the command that it should stop interpreting options, but echo will only output those:

$ echo -- -e
-- -e

You can use -e itself to get around the problem:

$ echo -e '\055e'
-e

Also, as others have pointed out, if you don't insist on using the bash builtin echo, your /bin/echo binary might be the GNU version of the tool (check the man page) and thus understand the POSIXLY_CORRECT environment variable:

$ POSIXLY_CORRECT=1 /bin/echo -e
-e

edited Dec 25 '08 at 21:13

answered Nov 26 '08 at 11:08

Do you happen to know, why `echo -e` does not work from Makefiles? – dma_k Oct 17 '10 at 00:00
2

@dma_k: works for me(tm). make(1) does not use the $SHELL builtin you are using, so it will behave differently. – Oct 18 '10 at 19:12
Thanks, great! Indeed, setting `SHELL=/bin/bash` in Makefile solved the problem! – dma_k Oct 22 '10 at 09:29
What about `/bin/echo -n`? POSIXLY_CORECT=1 won't work for it :( – Radu May 02 '15 at 20:06
1

@Radu: the other workaround is still available to you – May 03 '15 at 19:41

score 5 · Answer 3 · answered Nov 26 '08 at 11:12

5

There may be a better way, but this works:

printf -- "-e\n"

answered Nov 26 '08 at 11:12

Stephen Darlington

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In this case it works but it doesn't solve the global problem. Consider: printf "--version" – wheleph Nov 26 '08 at 12:25
2

@wheleph You have ignored the "--" before the "-e\n". This separates options from arguments. This is the best, portable (across shells) answer. +1. – richq Nov 26 '08 at 13:03
I agree that `printf` is the best, portable answer -- but disagree (strongly!) that embedding data in the format string is the right way to teach its use; that way lies people writing `printf -- "$var\n"` when they should be using `printf '%s\n' "$var"`, and that way lies format string injections (which, while not necessarily security-impacting in bash, can certainly lead to annoying bugs). – Charles Duffy Nov 16 '14 at 16:48

score 4 · Answer 4 · edited Oct 18 '12 at 15:18

4

You could cheat by doing

echo "-e "

That would be dash, e, space.

Alternatively you can use the more complex, but more precise:

echo -e \\\\x2De

edited Oct 18 '12 at 15:18

the Tin Man

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answered Nov 26 '08 at 11:06

Joachim Sauer

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score 1 · Answer 5 · answered Jul 08 '13 at 12:26

1

I like that one using a herestring:

cat <<<"-e"

answered Jul 08 '13 at 12:26

Tilman Vogel

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score 1 · Answer 6 · answered Nov 26 '08 at 11:14

1

 
[root@scintia mail]# POSIXLY_CORRECT=1; export POSIXLY_CORRECT
[root@scintia mail]# /bin/echo "-e"
-e
[root@scintia mail]#

answered Nov 26 '08 at 11:14

azerole

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Just 'POSIXLY_CORRECT=1 /bin/echo -e' (without the 's) is enough. Although this is not using the BASH version of echo – Vinko Vrsalovic Nov 26 '08 at 11:16
Information about Bash POSIX mode: http://www.network-theory.co.uk/docs/bashref/BashPOSIXMode.html. Item 41 addresses echo command. The only question left for me is why [root@scintia mail]# POSIXLY_CORRECT=1; export POSIXLY_CORRECT; echo "-e"; doesn't work – wheleph Nov 26 '08 at 16:06
Vinko, '/bin/echo -e' is also enough:) – wheleph Nov 26 '08 at 16:14

score 1 · Answer 7 · edited Oct 18 '12 at 15:18

1

After paying careful attention to the man page :)

SYSV3=1 /usr/bin/echo -e

works, on Solaris at least

edited Oct 18 '12 at 15:18

the Tin Man

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answered Nov 26 '08 at 11:44

The Archetypal Paul

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I haven't found any references to SYSV3 env variable in Linux echo man page – wheleph Nov 26 '08 at 16:12
Looks like the POSIXLY_CORRECT env var from another answer is the equivalent? – The Archetypal Paul Nov 27 '08 at 14:23

score 1 · Answer 8 · answered Nov 29 '08 at 22:38

1

Another alternative:

echo x-e | sed 's/^x//'

This is the way recommended by the autoconf manual:

[...] It is often possible to avoid this problem using 'echo "x$word"', taking the 'x' into account later in the pipe.

answered Nov 29 '08 at 22:38

CesarB

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score 0 · Answer 9 · answered Nov 26 '08 at 11:25

0

Another way:

echo -e' '
echo -e " \b-e"

answered Nov 26 '08 at 11:25

FerranB

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tho it outputs more than just "-e" under the hood :) – Johannes Schaub - litb Nov 26 '08 at 11:52

wheleph · Answer 10 · 2008-12-11T08:26:05.703

-2

/bin/echo -e

works, but why?

[resin@nevada ~]$ which echo 
/bin/echo

edited Dec 11 '08 at 08:26

answered Nov 26 '08 at 16:10

wheleph

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If you write "echo -e" it uses the bash internal echo, and writing "/bin/echo -e" uses the external /bin/echo command. But on my Ubuntu 8.04 box none of the two versions work.. – Anders Westrup Dec 11 '08 at 08:48
2

`which` will tell you which binary will be used, but it won't tell if your shell has a built-in function of the same name (which would probably be used instead of the binary). You can use `type` instead of `which` to find out more information. – dreamlax Apr 06 '10 at 04:48

echo "-e" doesn't print anything

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