How can I execute Perl code stored inside a shell script variable?

Question

I've got a script that calls Perl's Time::HiRes module to calculate elapsed time. Basically the script gets the time by passing the following one-liner: ``` use Time::HiRes qw(time); print time ``` to the Perl interpreter via backticks and gets back the results. ``` #/bin/sh START_TIME=`perl -e 'use Time::HiRes qw(time); print time'` END_TIME=`perl -e 'use Time::HiRes qw(time); print time'` ELAPSED_TIME=$(echo "($END_TIME - $START_TIME)" | bc) echo $ELAPSED_TIME ``` I tried to rewrite it in a more modular way, but I'm stumped by the quoting rules of the Bash shell. ``` #/bin/sh CALCULATE='bc' NOW="perl -e 'use Time::HiRes qw(time); print time'" START_TIME=`$NOW` [Some long running task ...] ELAPSED_TIME=$(echo "($NOW - $START_TIME)" | $CALCULATE) echo $ELAPSED_TIME ``` Bash complains that something is not quoted properly. Why doesn't Bash just expand the command in $NOW and pass it to the backtick to be executed? I tried various ways to embed Perl code in a shell script variable, but…

Accepted Answer

Using a function is the most straightforward way to do this, I think:

```
#! /bin/bash

now() {
    perl -e 'use Time::HiRes qw(time); print time';
}

calc=bc
time1=$(now)
time2=$(now)
elapsed=$(echo $time2 - $time1 | $calc)
echo $elapsed $time1 $time2
```

Essentially no quoting is required.

How can I execute Perl code stored inside a shell script variable?

Solution

Problem