in reply to RE: How Does Interpolation Work?
in thread How Does Interpolation Work?

I read your example, and wondered how this would interpolate:

my $R="foo";
print "${R}[0]\n";

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I thought it would either interpolate $R and print foo[0], or interpolate as $R[0], and print an empty string. Instead, I got an unexpected result:

[0]

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Anyone know why? I have a feeling I might be missing something in the meaning of ${R}[0].

These act in the expected way:

my $R = "foo";
my $R[0] = "bar";
print "${R}\[0]\n";
print "${R[0]}\n";

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Any enlightenment?

Alan

Update: Ah, that makes sense, a bit too subtle for me to get without some help. Thanks!

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RE: RE: How Does Interpolation Work?
by Russ (Deacon) on Jul 21, 2000 at 08:12 UTC
    Okay, you made $R lexical with my. ${R} is a symbolic reference to the scalar variable named 'R'. Just left alone, Perl will figure out you want the lexical variable and not a variable in the symbol table.

    Looking for ${R}[0], however, perl -w says this:
    Name "main::R" used only once: possible typo at - line 2.

    So, when you add characters that look like a variable specification Perl is only looking in the symbol table.

    If you don't make $R lexical, it will work like you originally expected: 

    $R="foo";
print "${R}[0]\n";

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    prints "foo[0]"

    To get perl to keep interpolating, just keep adding {}'s: 

    use vars qw/$R @foo/;
$R="foo";
@foo = (42);
print "${${R}}[0]\n";

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    prints 42

    :-)

    BTW, I seem remember some code posted some time ago which did this kind of "multiple interpolation." I can't find it. Anyone know what node that is (or am I just hallucinating again?
    ("Too much LDS" -- Kirk about Spock in ST:IV))

    Update: Here it is: Double Interpolation of a String

    Russ
    Brainbench 'Most Valuable Professional' for Perl

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