http://qs1969.pair.com?node_id=118046

ybiC has asked for the wisdom of the Perl Monks concerning the following question:

The following snippet is intended to iterate through my Cisco LAN switches to do stuff with each one.   My naming convention is switchNNNa, where NNN is a 3digit number greater than 100 and the last character is always the letter a.

But the "a" following the device number is tripping me up.   Perl thinks I mean "$ia" instead of "$i"a with this error: Global symbol "$ia" requires explicit package name at line 5

Suggestions from Wiser Monks?   Perl 5.00503 on Debian 2.3r3, in case it matters.
    cheers,
    Don
    striving toward Perl Adept
    (it's pronounced "why-bick") 

#!/usr/bin/perl -w
use strict;
my $i = 100;
for ($i..$i+200) {
   my $host = "switch$ia";
   print "$host\n";
   $i++;
}

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Replies are listed 'Best First'.
Re: "switch$ia" but don't interpolate trailing "a"
by jeroenes (Priest) on Oct 10, 2001 at 21:01 UTC
    Not that I am wiser, but: 
      my $host = "switch${i}a";

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    does the trick.
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(tye)Re: "switch$ia" but don't interpolate trailing "a"
by tye (Sage) on Oct 10, 2001 at 21:44 UTC

    My favorite is "switch$i$,a". (:

            - tye (but my friends call me "$,T$,y$,e$,")
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Re: "switch$ia" (Russ: Useless Benchmarking again)
by Russ (Deacon) on Oct 10, 2001 at 21:16 UTC
    I have to assume someone has done this before, but I didn't see it, so...

    Benchmarking interpolated strings vs. appended strings: 

    use Benchmark;

timethese( 1000000, {
  interp => sub {my $R = 1; my $S = "Test${R}a"},
  append => sub {my $R = 1; my $S = 'Test' . $R . 'a'}
});

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    results in: 
    Benchmark: timing 1000000 iterations of interp, manual...
  interp:  4 wallclock secs ( 3.68 usr +  0.05 sys =  3.73 CPU) @ 2680
+96.51/s
  manual:  3 wallclock secs ( 3.79 usr + -0.13 sys =  3.66 CPU) @ 2732
+24.04/s

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    Slightly faster to append strings together than to interpolate them, but statistically, it's a wash...

    Russ
    Brainbench 'Most Valuable Professional' for Perl

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      Someone please correct me if I'm wrong, ISTR interpolated strings are converted to concatenation at compile time?
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        You are correct, at least according to the B::Deparse documentation. Interpolation is just syntactic sugar for concatenation. You can see it for yourself: 

        $ perl -MO=Deparse,-p,-q -e '$foo = "BAR"; $bar = "a${foo}c"; $baz="a"
+.$foo."c"'                        
($foo = 'BAR');
($bar = (('a' . $foo) . 'c'));
($baz = (('a' . $foo) . 'c'));

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        Update: Added concatenation to the example for comparison. Also, the output using B::Terse to dump the opcode tree may be more convincing.

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        <sheepish grin>
        hehehe You're probably right.

        So, in the spirit of further useless benchmarking: 

        use Benchmark;

timethese( 10000000, {
  5 => sub {my $R = 5;},
  6 => sub {my $R = 6;}
});

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        results in: 
            5:  6 wallclock secs ( 6.61 usr + -0.03 sys =  6.58 CPU) @ 1519756
+.84/s
    6:  6 wallclock secs ( 5.89 usr +  0.08 sys =  5.97 CPU) @ 1675041
+.88/s

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        As you can see, it is slightly faster to assign a 6 than a 5, so I will use 6's more often from now on... :-)
        </sheepish grin>

        Russ
        Brainbench 'Most Valuable Professional' for Perl

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