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I'm trying to assign -e to a variable in Bash 4. But the variable remains empty.

For example:

$ var="-e" 
$ echo $var

$ var='-e' 
$ echo $var

$ var="-t" 
$ echo $var
-t

Why does it work with -t, but not -e?

Jeff Schaller
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Dude
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    Outputting the variable with echo would output nothing, but the variable has the value -e (which is a valid option for echo in bash). You'd have the same issue with -n and -E and combinations thereof. – Kusalananda Aug 25 '19 at 07:43

2 Answers2

5

It works, but running echo -e doesn't output anything in Bash unless both the posix and xpg_echo options are enabled, as the -e is then interpreted as an option:

$ help echo
echo: echo [-neE] [arg ...]
    Write arguments to the standard output.

    Display the ARGs, separated by a single space character and followed by a
    newline, on the standard output.

    Options:
      -n        do not append a newline
      -e        enable interpretation of the following backslash escapes
      -E        explicitly suppress interpretation of backslash escapes

Use 

printf "%s\n" "$var"

instead.

And as cas notes in a comment, Bash's declare -p var (typeset -p var in ksh, zsh and yash (and bash)) can be used to tell the exact type and contents of a variable.

See: Why is printf better than echo?

Stéphane Chazelas
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ilkkachu
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    also, if the OP wants to verify that $var does indeed contain -e, they can run declare -p var. – cas Aug 25 '19 at 07:54
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A command (like your echo) takes command line arguments which could be flags (-e in your case). Many commands (at least common Linux versions) understand -- (two hyphens) as "end of flags, whatever follows is regular arguments". So you can delete a file perversely named -r by rm -- -r.

For displaying stuff, printf is more robust all around (if harder to use).

vonbrand
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