Command substitution in quotes cannot be used to call the variable

Question

Say:

variable="Something that it holds"

then echo "$variable" will output: Something that it holds

But say I also do:

var2="variable";  
echo "\$$(echo $var2)"

will just output: $variable
And not: Something that it holds

Can anyone tell me about what feature of Unix is in play, here?

Answer 1

variable="Something"
var2="variable";
echo "\$$(echo $var2)"

In the last line, you expect "\$$(echo $var2)" → $variable → Something. This would require the shell to perform two parameter expansions. It does not, but rather simply prints the result of the command substitution $(echo $var2) prepended by a $.

In principle, eval helps you to get what you want. After the shell performs the first step, "\$$(echo $var2)" → $variable, eval performs the second step, $variable → Something.

$ eval echo "\$$(echo $var2)"
Something

Although in our particular case the above command is OK, that still lacks a correct quoting, and printf is to be favored over echo,

$ eval 'printf "%s\n" "${'"$var2"'}"'
Something

However, eval raises security concerns with untrusted data. Suppose var2="variable;rm importantFile". In that case, eval passes

echo $variable;rm importantFile

to the shell, which happily removes importantFile, if it exists.

In some shells (e.g.: Bash, ksh, Zsh) you can also do it using indirection. The syntax of indirect expansion in Bash is:

$ echo "${!var2}"
Something

var2="variable;rm importantFile" is not a problem anymore, but var2='a[$(rm importantFile)]' still is.

Read more about indirection in the Bash manual.

If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter

Answer 2

Smells like an anti-pattern(*). Many shells have string-indexed arrays/dictionaries. In ksh93 (where that syntax comes from), bash or zsh:

typeset -A dictionary
x="keyX"
y="keyY"
dictionary[keyX]="valueX"
dictionary[$y]="valueY"

printf '%s\n' "${dictionary[$x]}"
printf '%s\n' "${dictionary[keyY]}"

(*) Nothing to do with Linux per se. The variable-name-in-a-variable is a very general "anti-pattern", something that shouldn't be done, and if you think you need it, there is a problem with your design (XY problem)? Most uses of a variable-name-in-a-variable are better replaced with a dictionary when that exists.

Answer 3

What you are observing is the standard behavior of a POSIX shell: in general, it

1. reads its input;

2. breaks the input into tokens: words and operators (token recognition);

   • during this step, any time it encounters an unquoted $ (or `), it recursively determines the type of expansion and the token to be expanded, reading all the needed input;

3. parses the input into simple commands and compound commands;

4. performs various expansions (separately) on different parts of each command, resulting in a list of pathnames and fields to be treated as a command and arguments;

5. performs redirection and removes redirection operators and their operands from the parameter list;

6. executes a function, built-in, executable file, or script;

7. optionally waits for the command to complete and collects the exit status.

When parsing echo "\$$(echo $var2)", the shell detects two expansions (step 2): the double-quoted command substitution $(echo $var2) and the unquoted parameter expansion $var2. The escaped $ in \$ is taken as a literal dollar sign because a double-quoted \ retains its role as an escape character when followed by $.

No further detection of expansions happens at later stages. Specifically, there is no further parsing of the result of expansions performed in step 4 ("\$$(echo $var2)" → "\$$(echo variable)" → "\$variable" → $variable) that could detect expansion-triggering characters.

Also note that, while the $ symbol is used to replace the name of a variable with its content in the context of parameter expansion, it has not been designed as a general dereference operator.

In standard parameter expansion, whose simplest form is ${parameter}, the parameter specification is only allowed to be a variable name, a positional parameter or a special parameter (see the definition of "parameter"). Strictly speaking, parameter expansions can not be nested (an expansion expression is only allowed as word in the various ${parameter<symbols>[word]} forms).

You can easily verify that, with the exception of the Z shell, ${${foo}} is not a valid expression and that $$ expands to the shell's PID (thus, $foo expands to the value of foo, $$foo expands to the concatenation of the shell's PID and the literal "foo", $$$foo expands to the concatenation of the shell's PID and the value of foo, ...).

Answer 4

In addition to indirect variable expansion, in bash (starting from version 4.3) you can use a nameref

declare -n var2="variable"  # "var2" is a _reference_ to "variable"

variable="Something"
echo "$var2"     # => Something

variable="something else"
echo "$var2"     # => something else

unset variable
echo "$var2"     # => ""

---

I don't know how this is implemented, but this is an interesting tidbit: you can find out what a nameref refers to using indirection:

echo ${!var2}    # => variable