why this works? quoted strings

patcat88 has asked for the wisdom of the Perl Monks concerning the following question:

#!/usr/bin/perl -w
$ALIAS = 0;
$arg = 'ST(0)';
$var = 'self';
$pname = 'MyMod::MyFunc';
$expr = '\n    if (SvROK($arg) && SvTYPE(SvRV($arg))==SVt_PVHV)
        $var = (HV*)SvRV($arg);
    else
        Perl_croak(aTHX_ \"%s: %s is not a hash reference\",
            ${$ALIAS?\q[GvNAME(CvGV(cv))]:\qq[\"$pname\"]},
            \"$var\");\n'
;
 $str = qq/"\\n$expr;\\n"/;
$res = eval $str;
0;
[download]

Can someone explain why this works? All of it. Its from ParseXS. I also can't find any documentation on "${" operator.

edit: changed typemap string from double to single quotes so eval works

Comment on why this works? quoted strings Download Code

Replies are listed 'Best First'.
Re: why this works? quoted strings by moritz (Cardinal) on Oct 25, 2011 at 12:30 UTC
`${ ... }` just interpolates the result from an expression, in this case ($ALIAS being 0) it is `\"$pname\"`. With the example you gave, the eval fails, so I don't know what's the point of it. Perl 6 - second systems done right	[reply] [d/l] [select]
Re^2: why this works? quoted strings by patcat88 (Deacon) on Oct 25, 2011 at 12:49 UTC
The eval came from dist/ExtUtils-ParseXS/lib/ExtUtils/ParseXS.pm#l1884 in perl.git Yep, the eval doesn't work. I shouldve used single quotes, not double for the typemap string.	[reply]
Re^2: why this works? quoted strings by ikegami (Patriarch) on Oct 25, 2011 at 21:19 UTC
`$...` interpolates. Except for `${BAREWORD}` (which means `$BAREWORD`), `${ ... }` dereferences first. Just like outside of string literals.	[reply] [d/l] [select]
Re: why this works? quoted strings by mrstlee (Beadle) on Oct 25, 2011 at 16:07 UTC
${ isn't an operator itself - it has to be book-ended with a }. Basically you can put any legal perl inside the {}'s - so long as it evaluates to a reference. You then de-reference according to the type. It makes for some interesting strings: `print "My scalar = ${\(my $scalar = 5)}\n"; print "My array = @{[(0..9)]}\n"; print "My array via scalar references passed to map = @{[ map {$$_} \( +0..9) ]}\n"; print "My array returned by anonymous subroutine = @{[do {(0..9)}]}\n" +; print "My array formatted with join = ${\(join '',(0..9))}\n"; print "My hash treated as array = @{ [my %h = (0..9)] }\n";` [download] Prints: `My scalar = 5 My array = 0 1 2 3 4 5 6 7 8 9 My array via scalar references passed to map = 0 1 2 3 4 5 6 7 8 9 My array returned by anonymous subroutine = 0 1 2 3 4 5 6 7 8 9 My array formatted with join = 0123456789 My hash treated as array = 0 1 2 3 4 5 6 7 8 9` [download]	[reply] [d/l] [select]


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