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").

Answer 1

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

Answer 2

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

Answer 3

There may be a better way, but this works:

printf -- "-e\n"

Answer 4

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

Answer 5

Another way:

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

Answer 6

After paying careful attention to the man page :)

SYSV3=1 /usr/bin/echo -e

works, on Solaris at least

Paul.

Answer 7

/bin/echo -e

works, but why?

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

Answer 8

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.

Answer 9

The one true way to print any arbitrary string:

printf "%s" "$vars"