Dereferencing sh variables

[J65nko]

In a Linux Journal Tech tip is shown how you can dereference a variable, when it is passed as string to a function. It uses the bash ! operator.

DerefernceVariablePassedToFunction() {
    if [ -n "$1" ] ; then
        echo "value of [${1}] is: [${!1}]"
    else
        echo "Null parameter passed to this function"
    fi
}

Variable="LinuxJournal"
DerefernceVariablePassedToFunction Variable

This will now print LinuxJournal.

I just wondered why he did not use eval like this

#/bin/sh

dereference() {
    if [ -n "$1" ] ; then
        eval echo value of \$1 is: \$$1
    else
        echo "Null parameter passed to this function"
    fi
}

variable="Just use 'eval'"
dereference variable

This has the same effect. A run with sh -vx so you can see the double interpretation pass performed by eval

 sh -vx dereference.sh 
#/bin/sh

dereference() {
    if [ -n "$1" ] ; then
        eval echo value of \$1 is: \$$1
    else
        echo "Null parameter passed to this function"
    fi
}

variable="Just use 'eval'"
+ variable=Just use 'eval'
dereference variable 
+ dereference variable
echo value of $1 is: $variable
value of variable is: Just use 'eval'

The first pass converts

echo value of \$1 is: \$$1

into

echo value of $1 is: $variable

\$1 is seen by the shell as a "$" followed by a '1' . The "\" prevents
the shell from interpreting $1 as the first parameter, as it normally would do.

\$$1is parsed as a literal "$" followed by the function parameter $1 which contains the string "variable"


The second and last pass re-processes this

echo value of $1 is: $variable

and without any "\" to prevent variable expansion, produces the wanted result:

value of variable is: Just use 'eval'