01-17-2008
scalar variable assignment in perl + { operator
When reading over some perl code in a software document, I came across an assignment statement like this
$PATH = ${PROJECT}/......./....
In this particular form of scalar variable assignment, what does the curly braces operators do ? Also, what is the benefit in doing scalar assignment this way ?
many thanks
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LEARN ABOUT SUSE
glib::flags
Glib::Flags(3) User Contributed Perl Documentation Glib::Flags(3)
NAME
Glib::Flags
DESCRIPTION
Glib maps flag and enum values to the nicknames strings provided by the underlying C libraries. Representing flags this way in Perl is an
interesting problem, which Glib solves by using some cool overloaded operators.
The functions described here actually do the work of those overloaded operators. See the description of the flags operators in the "This
Is Now That" section of Glib for more info.
HIERARCHY
Glib::Flags
METHODS
scalar = $class->new ($a)
o $a (scalar)
Create a new flags object with given bits. This is for use from a subclass, it's not possible to create a "Glib::Flags" object as such.
For example,
my $f1 = Glib::ParamFlags->new ('readable');
my $f2 = Glib::ParamFlags->new (['readable','writable']);
An object like this can then be used with the overloaded operators.
scalar = $a->all ($b, $swap)
o $b (scalar)
o $swap (scalar)
ref = $a->as_arrayref
integer = $a->bool ($b, $swap)
o $b (scalar)
o $swap (integer)
integer = $a->eq ($b, $swap)
o $b (scalar)
o $swap (integer)
integer = $a->ge ($b, $swap)
o $b (scalar)
o $swap (integer)
scalar = $a->intersect ($b, $swap)
o $b (scalar)
o $swap (scalar)
integer = $a->ne ($b, $swap)
o $b (scalar)
o $swap (integer)
scalar = $a->sub ($b, $swap)
o $b (scalar)
o $swap (scalar)
scalar = $a->union ($b, $swap)
o $b (scalar)
o $swap (scalar)
scalar = $a->xor ($b, $swap)
o $b (scalar)
o $swap (scalar)
SEE ALSO
Glib
COPYRIGHT
Copyright (C) 2003-2009 by the gtk2-perl team.
This software is licensed under the LGPL. See Glib for a full notice.
perl v5.12.1 2010-07-05 Glib::Flags(3)