11.5. Taking References to Scalars11.5.2. SolutionTo create a reference to a scalar variable, use the backslash operator: $scalar_ref = \$scalar; # get reference to named scalar To create a reference to an anonymous scalar value (a value that isn't in a variable), assign to a dereferenced undefined variable: undef $anon_scalar_ref; $$anon_scalar_ref = 15; This creates a reference to a constant scalar: $anon_scalar_ref = \15; Use ${...} to dereference: print ${ $scalar_ref }; # dereference it ${ $scalar_ref } .= "string"; # alter referent's value 11.5.3. DiscussionIf you want to create many new anonymous scalars, use a subroutine that returns a reference to a lexical variable out of scope, as explained in this chapter's Introduction: sub new_anon_scalar { my $temp; return \$temp; } Dereference a scalar reference by prefacing it with $ to get at its contents: $sref = new_anon_scalar( ); $$sref = 3; print "Three = $$sref\n"; @array_of_srefs = ( new_anon_scalar( ), new_anon_scalar( ) ); ${ $array[0] } = 6.02e23; ${ $array[1] } = "avocado"; print "\@array contains: ", join(", ", map { $$_ } @array ), "\n"; Notice we put braces around $array[0] and $array[1]. If we tried to say $$array[0], the tight binding of dereferencing would turn it into $array->[0]. It would treat $array as an array reference and return the element at index zero. Here are other examples where it is safe to omit the braces: $var = `uptime`; # $var holds text $vref = \$var; # $vref "points to" $var if ($$vref =~ /load/) { } # look at $var, indirectly chomp $$vref; # alter $var, indirectly As mentioned in the Introduction, you may use the ref built-in to inspect a reference for its referent's type. Calling ref on a scalar reference returns the string "SCALAR": # check whether $someref contains a simple scalar reference if (ref($someref) ne "SCALAR") { die "Expected a scalar reference, not $someref\n"; } 11.5.4. See AlsoChapters 8 and 9 of Programming Perl and perlref(1) Copyright © 2003 O'Reilly & Associates. All rights reserved. |
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