sort - sort a list of values
sort SUBNAME LIST
sort BLOCK LIST
sort LIST
Sorts the
LIST and returns the sorted list value. If
SUBNAME or
BLOCK
is omitted, sorts in standard string comparison order. If
SUBNAME is
specified, it gives the name of a subroutine that returns an integer
less than, equal to, or greater than 0, depending on how the elements
of the list are to be ordered. (The <=> and cmp
operators are extremely useful in such routines.)
SUBNAME may be a
scalar variable name (unsubscripted), in which case the value provides
the name of (or a reference to) the actual subroutine to use. In place
of a
SUBNAME, you can provide a
BLOCK as an anonymous, in-line sort
subroutine.
If the subroutine's prototype is ($$), the elements to be compared
are passed by reference in @_, as for a normal subroutine. This is
slower than unprototyped subroutines, where the elements to be
compared are passed into the subroutine
as the package global variables $a and $b (see example below). Note that
in the latter case, it is usually counter-productive to declare $a and
$b as lexicals.
In either case, the subroutine may not be recursive. The values to be compared are always passed by reference, so don't modify them.
You also cannot exit out of the sort block or subroutine using any of the
loop control operators described in the perlsyn manpage or with goto.
When use locale is in effect, sort LIST sorts
LIST according to the
current collation locale. See the perllocale manpage.
Examples:
# sort lexically
@articles = sort @files;
# same thing, but with explicit sort routine
@articles = sort {$a cmp $b} @files;
# now case-insensitively
@articles = sort {uc($a) cmp uc($b)} @files;
# same thing in reversed order
@articles = sort {$b cmp $a} @files;
# sort numerically ascending
@articles = sort {$a <=> $b} @files;
# sort numerically descending
@articles = sort {$b <=> $a} @files;
# this sorts the %age hash by value instead of key
# using an in-line function
@eldest = sort { $age{$b} <=> $age{$a} } keys %age;
# sort using explicit subroutine name
sub byage {
$age{$a} <=> $age{$b}; # presuming numeric
}
@sortedclass = sort byage @class;
sub backwards { $b cmp $a }
@harry = qw(dog cat x Cain Abel);
@george = qw(gone chased yz Punished Axed);
print sort @harry;
# prints AbelCaincatdogx
print sort backwards @harry;
# prints xdogcatCainAbel
print sort @george, 'to', @harry;
# prints AbelAxedCainPunishedcatchaseddoggonetoxyz
# inefficiently sort by descending numeric compare using
# the first integer after the first = sign, or the
# whole record case-insensitively otherwise
@new = sort {
($b =~ /=(\d+)/)[0] <=> ($a =~ /=(\d+)/)[0]
||
uc($a) cmp uc($b)
} @old;
# same thing, but much more efficiently;
# we'll build auxiliary indices instead
# for speed
@nums = @caps = ();
for (@old) {
push @nums, /=(\d+)/;
push @caps, uc($_);
}
@new = @old[ sort {
$nums[$b] <=> $nums[$a]
||
$caps[$a] cmp $caps[$b]
} 0..$#old
];
# same thing, but without any temps
@new = map { $_->[0] }
sort { $b->[1] <=> $a->[1]
||
$a->[2] cmp $b->[2]
} map { [$_, /=(\d+)/, uc($_)] } @old;
# using a prototype allows you to use any comparison subroutine
# as a sort subroutine (including other package's subroutines)
package other;
sub backwards ($$) { $_[1] cmp $_[0]; } # $a and $b are not set here
package main;
@new = sort other::backwards @old;
If you're using strict, you must not declare $a
and $b as lexicals. They are package globals. That means
if you're in the main package, it's
@articles = sort {$main::b <=> $main::a} @files;
or just
@articles = sort {$::b <=> $::a} @files;
but if you're in the FooPack package, it's
@articles = sort {$FooPack::b <=> $FooPack::a} @files;
The comparison function is required to behave. If it returns
inconsistent results (sometimes saying $x[1] is less than $x[2] and
sometimes saying the opposite, for example) the results are not
well-defined.