in reply to Resolving scalars in a scalar

I arrive too late and Zaxo has already posted a solution similar to mine. So instead of repeating the same thing, I have considerred Zaxo's solution and came up with an idea, this may be insane but here it goes: tokenize the input string and evaluate all perl variables, sort of a mini compiler/preprocessor. The tokeniser will be very difficult to implement, I have considered B, but got lazy. Anyway, below is my solution that will work under very specific conditions:
use strict;

my $alpha = "A";
my $delta = "D";
my $theta = "T";
my @hash = qw/ Hello world /;

my $test = '@hash $delta and $alpha or $theta are good';

# evaluate perl variables except bare words
my $result = qq{@{[ map { m/^[\$@%]./ ? eval $_ : $_ }
                        split/\s/,$test ]}};
print "$result\n";

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And the output - 
Hello world D and A or T are good

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Re: Re: Resolving scalars in a scalar
by SavannahLion (Pilgrim) on Oct 30, 2003 at 08:35 UTC
    So after mindlessly fiddling with the eval statement, I came up with the following; 
    my $alpha = "A";
my $delta = "D";
my $test = "$delta and $alpha";
eval "\$test = \"$test\"";
print "$test";

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    Is that... right?

    Is it fair to stick a link to my site here?

    Thanks for you patience.

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[d/l]
      Ummm, because your $test is quoted inside double quote, your $test = "D and A" even before you go into the eval, so the eval had no effect.

      Change the code slightly as follows -
      my $alpha = "A";
my $delta = "D";
my $test = '$delta and $alpha';
eval "\$test = \"$test\"";
print "$test";

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      and yes, it will work this time, because the code is equivalent to -
      my $alpha = "A";
my $delta = "D";
my $test = '$delta and $alpha';
$test = "$delta and $alpha";
print "$test";

      [download]
      Where the variable $test inside double quote " " has been expanded by the perl interpreter to the value of $test before going into eval

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