Bash and Zsh – Double and Triple Variable Substitution

bashvariable substitutionzsh

Follow-up to the background part in this question.

In bash I can use ${!FOO} for double substitution, in zsh ${(P)FOO}. In both, the old-school (hack-y) eval \$$FOO works.

So, the smartest and most logical thing for me would be ${${FOO}}, ${${${FOO}}}… for double/triple/n substitution. Why doesn’t this work as expected?

Second: What does the \ do in the eval statement? I reckon it’s an escape, making something like eval \$$$FOO impossible. How to do a triple/n substitution with that that works in every shell?

Best Answer

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

Related Solutions

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

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.)

Bash – Escaping strings in associative arrays (bash)

First, create an array with the parameters.

Then, store the array string value (found with declare -p) at parameters, and recover and use it as an actual array later on like in:

#!/bin/bash

programs=( "ls" "echo" )
declare -A parameters

arrayTmp=("-l" "/tmp/foo bar")
parameters["ls"]="`declare -p arrayTmp |sed -r "s,[^=]*='(.*)'$,\1,"`"

parameters["echo"]="Hello"

for program in "${programs[@]}";do
    echo "PROGRAM: $program"
    declare -a arrayTmp="${parameters[$program]}"
    $program "${arrayTmp[@]}"
    arrayTmp=()
done

Best Answer

Related Solutions

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

Bash – Escaping strings in associative arrays (bash)

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