I just assigned a variable, but echo $variable shows something else

You may want to know why this is happening. Together with the great explanation by that other guy, find a reference of Why does my shell script choke on whitespace or other special characters? written by Gilles in Unix & Linux:

Why do I need to write "$foo"? What happens without the quotes?

$foo does not mean “take the value of the variable foo”. It means something much more complex:

• First, take the value of the variable.
• Field splitting: treat that value as a whitespace-separated list of fields, and build the resulting list. For example, if the variable contains foo * bar ​ then the result of this step is the 3-element list foo, *, bar.
• Filename generation: treat each field as a glob, i.e. as a wildcard pattern, and replace it by the list of file names that match this pattern. If the pattern doesn't match any files, it is left unmodified. In our example, this results in the list containing foo, following by the list of files in the current directory, and finally bar. If the current directory is empty, the result is foo, *, bar.

Note that the result is a list of strings. There are two contexts in shell syntax: list context and string context. Field splitting and filename generation only happen in list context, but that's most of the time. Double quotes delimit a string context: the whole double-quoted string is a single string, not to be split. (Exception: "$@" to expand to the list of positional parameters, e.g. "$@" is equivalent to "$1" "$2" "$3" if there are three positional parameters. See What is the difference between $* and $@?)

The same happens to command substitution with $(foo) or with `foo`. On a side note, don't use `foo`: its quoting rules are weird and non-portable, and all modern shells support $(foo) which is absolutely equivalent except for having intuitive quoting rules.

The output of arithmetic substitution also undergoes the same expansions, but that isn't normally a concern as it only contains non-expandable characters (assuming IFS doesn't contain digits or -).

See When is double-quoting necessary? for more details about the cases when you can leave out the quotes.

Unless you mean for all this rigmarole to happen, just remember to always use double quotes around variable and command substitutions. Do take care: leaving out the quotes can lead not just to errors but to security holes.

In all of the cases above, the variable is correctly set, but not correctly read! The right way is to use double quotes when referencing:

echo "$var"

This gives the expected value in all the examples given. Always quote variable references!

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Why?

When a variable is unquoted, it will:

1. Undergo field splitting where the value is split into multiple words on whitespace (by default):

   Before: /* Foobar is free software */

   After: /*, Foobar, is, free, software, */

2. Each of these words will undergo pathname expansion, where patterns are expanded into matching files:

   Before: /*

   After: /bin, /boot, /dev, /etc, /home, ...

3. Finally, all the arguments are passed to echo, which writes them out separated by single spaces, giving

   /bin /boot /dev /etc /home Foobar is free software Desktop/ Downloads/
   

   instead of the variable's value.

When the variable is quoted it will:

1. Be substituted for its value.
2. There is no step 2.

This is why you should always quote all variable references, unless you specifically require word splitting and pathname expansion. Tools like shellcheck are there to help, and will warn about missing quotes in all the cases above.

In addition to other issues caused by failing to quote, -n and -e can be consumed by echo as arguments. (Only the former is legal per the POSIX spec for echo, but several common implementations violate the spec and consume -e as well).

To avoid this, use printf instead of echo when details matter.

Thus:

$ vars="-e -n -a"
$ echo $vars      # breaks because -e and -n can be treated as arguments to echo
-a
$ echo "$vars"
-e -n -a

However, correct quoting won't always save you when using echo:

$ vars="-n"
$ echo "$vars"
$ ## not even an empty line was printed

...whereas it will save you with printf:

$ vars="-n"
$ printf '%s\n' "$vars"
-n