package overload;
+our $VERSION = '1.00';
+
$overload::hint_bits = 0x20000;
sub nil {}
binary => "& | ^",
unary => "neg ! ~",
mutators => '++ --',
- func => "atan2 cos sin exp abs log sqrt",
+ func => "atan2 cos sin exp abs log sqrt int",
conversion => 'bool "" 0+',
iterators => '<>',
dereferencing => '${} @{} %{} &{} *{}',
=item * I<Transcendental functions>
- "atan2", "cos", "sin", "exp", "abs", "log", "sqrt",
+ "atan2", "cos", "sin", "exp", "abs", "log", "sqrt", "int"
If C<abs> is unavailable, it can be autogenerated using methods
for "E<lt>" or "E<lt>=E<gt>" combined with either unary minus or subtraction.
+Note that traditionally the Perl function L<int> rounds to 0, thus for
+floating-point-like types one should follow the same semantic. If
+C<int> is unavailable, it can be autogenerated using the overloading of
+C<0+>.
+
=item * I<Boolean, string and numeric conversion>
"bool", "\"\"", "0+",
return any arbitrary Perl value. If the corresponding operation for this value
is overloaded too, that operation will be called again with this value.
+As a special case if the overload returns the object itself then it will
+be used directly. An overloaded conversion returning the object is
+probably a bug, because you're likely to get something that looks like
+C<YourPackage=HASH(0x8172b34)>.
+
=item * I<Iteration>
"<>"
Both these problems can be cured. Say, if we want to overload hash
dereference on a reference to an object which is I<implemented> as a
hash itself, the only problem one has to circumvent is how to access
-this I<actual> hash (as opposed to the I<virtual> exhibited by
+this I<actual> hash (as opposed to the I<virtual> hash exhibited by the
overloaded dereference operator). Here is one possible fetching routine:
sub access_hash {
$out;
}
-To move creation of the tied hash on each access, one may an extra
+To remove creation of the tied hash on each access, one may an extra
level of indirection which allows a non-circular structure of references:
package two_refs1;
$a->[$key];
}
-Now if $baz is overloaded like this, then C<$bar> is a reference to a
+Now if $baz is overloaded like this, then C<$baz> is a reference to a
reference to the intermediate array, which keeps a reference to an
actual array, and the access hash. The tie()ing object for the access
-hash is also a reference to a reference to the actual array, so
+hash is a reference to a reference to the actual array, so
=over
I<components> $a and $b of an object. In the above subroutine
C<"[$meth $a $b]"> is a catenation of some strings and components $a
and $b. If these components use overloading, the catenation operator
-will look for an overloaded operator C<.>, if not present, it will
+will look for an overloaded operator C<.>; if not present, it will
look for an overloaded operator C<"">. Thus it is enough to use
use overload nomethod => \&wrap, '""' => \&str;
(not required without mutators!), and implements only those arithmetic
operations which are used in the example.
-To implement most arithmetic operations is easy, one should just use
+To implement most arithmetic operations is easy; one should just use
the tables of operations, and change the code which fills %subr to
my %subr = ( 'n' => sub {$_[0]} );
way to know that the implementation of C<'+='> does not mutate
the argument, compare L<Copy Constructor>).
-To implement a copy constructor, add C<'=' => \&cpy> to C<use overload>
+To implement a copy constructor, add C<< '=' => \&cpy >> to C<use overload>
line, and code (this code assumes that mutators change things one level
deep only, so recursive copying is not needed):
https://perl5.git.perl.org / perl5.git / blobdiff