Why does the following snippet work at all? And what evil might be possible using this? But seriously, is there any reason, the code in ${} gets evaluated at al
From the "Using References" section of the perlref documentation:
Anywhere you'd put an identifier (or chain of identifiers) as part of a variable or subroutine name, you can replace the identifier with a BLOCK returning a reference of the correct type. In other words, the previous examples could be written like this:
$bar = ${$scalarref}; push(@{$arrayref}, $filename); ${$arrayref}[0] = "January"; ${$hashref}{"KEY"} = "VALUE"; &{$coderef}(1,2,3); $globref->print("output\n"); # iff IO::Handle is loadedAdmittedly, it's a little silly to use the curlies in this case, but the BLOCK can contain any arbitrary expression, in particular, subscripted expressions:
&{ $dispatch{$index} }(1,2,3); # call correct routineBecause of being able to omit the curlies for the simple case of
$$x, people often make the mistake of viewing the dereferencing symbols as proper operators, and wonder about their precedence. If they were, though, you could use parentheses instead of braces. That's not the case. Consider the difference below; case 0 is a short-hand version of case 1, not case 2:$$hashref{"KEY"} = "VALUE"; # CASE 0 ${$hashref}{"KEY"} = "VALUE"; # CASE 1 ${$hashref{"KEY"}} = "VALUE"; # CASE 2 ${$hashref->{"KEY"}} = "VALUE"; # CASE 3Case 2 is also deceptive in that you're accessing a variable called
%hashref, not dereferencing through$hashrefto the hash it's presumably referencing. That would be case 3.
Later in "Symbolic references":
We said that references spring into existence as necessary if they are undefined, but we didn't say what happens if a value used as a reference is already defined, but isn't a hard reference. If you use it as a reference, it'll be treated as a symbolic reference. That is, the value of the scalar is taken to be the name of a variable, rather than a direct link to a (possibly) anonymous value.