Bash and Zsh – Equivalent of Bash Indirect Referencing ${!FOO} in Zsh

bashforvariable substitutionzsh

${!FOO} performs a double substitution in bash, meaning it takes the (string) value of FOO and uses it as a variable name.
zsh doesn’t support this feature.

Is there a way to make this work the same in bash and zsh?

Background:

I’ve got a list of environment variables, like

PATH MAIL EDITOR

and want to first print the variable names and afterwards their values.

This works in bash but not zsh:

for VAR in LIST
do
        echo $VAR
        echo ${!VAR}
done

It should be somehow possible “the old way” with eval, but I can’t get it to work:

for VAR in LIST
do
        echo $VAR
        echo `eval \$$VAR`
done

I’m never going to understand why I can’t simply do arbitrary deep substitutions like ${${VAR}} or even ${${${VAR}}} if need be, so an explanation for that would be nice, too.

Best Answer

Both bash and zsh have a way to perform indirect expansion, but they use different syntax.

It's easy enough to perform indirect expansion using eval; this works in all POSIX and most Bourne shells. Take care to quote properly in case the value contains characters that have a special meaning in the shell.

eval "value=\"\${$VAR}\""
echo "$VAR"
echo "$value"

${${VAR}} doesn't work because it's not a feature that any shell implements. The thing inside the braces must conform to syntax rules which do not include ${VAR}. (In zsh, this is supported syntax, but does something different: nested substitutions perform successive transformations on the same value; ${${VAR}} is equivalent to $VAR since this performs the identity transformation twice on the value.)

Related Solutions

Bash and Zsh – Double and Triple Variable Substitution

The \ must be used to prevent the expansion of $$ (current process id). For triple substitution, you need double eval, so also more escapes to avoid the unwanted expansions in each eval:

#! /bin/bash
l0=value
l1=l0
l2=l1
l3=l2
l4=l3

echo $l0
eval echo \$$l1
eval eval echo \\$\$$l2
eval eval eval echo \\\\$\\$\$$l3
eval eval eval eval echo \\\\\\\\$\\\\$\\$\$$l4

The effect of “${(@f)…}” in Zsh

The character inside the parentheses are parameter expansion flags. They can be used around a variable substitution or a command substitution, e.g.

${(@f)SOME_VARIABLE}
${(@f)${(xyz)SOME_VARIABLE}}
${(@f)$(some-command)}

The flag f splits the result of the expansion at newlines. The flag @ ensures that the resulting array is expanded as multiple words; oddly, it only has effect inside double quotes, where it acts as a generalization of "$@".

Usually, these flags are used like this:

lines=("${(@f)$(egrep "$2" "$file")}")

This way lines becomes an array where each element is one line of output from egrep.

Here, the expansion is not in a context that allows multiple words, so the @ flag has no effect. The effect of the f flag is somewhat unintuitive: it forces an array context, so the IFS-separated words in the egrep output are stored in a temporary array which is then joined with the first character of IFS.

linestack=${(@f)$(egrep "$2" "$file")}

Contrast:

$ zsh -c 'a=${(f)$(echo hello world; echo wibble)}; print -lr $#a $a'
18
hello world wibble
$ zsh -c 'IFS=:$IFS; a=${(f)$(echo hello world; echo wibble)}; print -lr $#a $a'
18
hello:world:wibble
$ zsh -c 'IFS=:; a=${(f)$(echo hello world; echo wibble)}; print -lr $#a $a'
18
hello world
wibble

$ zsh -c 'a=(${(f)$(echo hello world; echo wibble)}); print -lr $#a $a' 
1
hello world wibble
$ zsh -c 'a=("${(@f)$(echo hello world; echo wibble)}"); print -lr $#a $a'
2
hello world
wibble

Background:

Best Answer

Related Solutions

Bash and Zsh – Double and Triple Variable Substitution

The effect of “${(@f)…}” in Zsh

Related Question