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by gods
on Aug 24, 1999 at 22:43 UTC ( #344=perlfunc: print w/ replies, xml ) Need Help??


See the current Perl documentation for sort.

Here is our local, out-dated (pre-5.6) version:

sort - sort a list of values



sort LIST

Sorts the LIST and returns the sorted list value. If SUBNAME or BLOCK is omitted, sort()s 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 array 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.

In the interests of efficiency the normal calling code for subroutines is bypassed, with the following effects: the subroutine may not be a recursive subroutine, and the two elements to be compared are passed into the subroutine not via @_ but as the package global variables $a and $b (see example below). They are passed by reference, so don't modify $a and $b. And don't try to declare them as lexicals either.

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.


    # 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;

    # sort using explicit subroutine name
    sub byage {
        $age{$a} <=> $age{$b};  # presuming numeric
    @sortedclass = sort byage @class;

    # this sorts the %age hash by value instead of key
    # using an in-line function
    @eldest = sort { $age{$b} <=> $age{$a} } keys %age;

    sub backwards { $b cmp $a; }
    @harry = ('dog','cat','x','Cain','Abel');
    @george = ('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 using a Schwartzian Transform (no temps)
    @new = map { $_->[0] }
        sort { $b->[1] <=> $a->[1]
               $a->[2] cmp $b->[2]
        } map { [$_, /=(\d+)/, uc($_)] } @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.

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