sub overlap{ 
    ($_[1] < $_[3] ? $_[1]: $_[3]) 
    - 
    ($_[0] > $_[2] ? $_[0] : $_[2]) + 1; 
}

@data = qw[1 5 2 6]; 
print "\n@data : ", overlap @data 
    and @data = map($_-1, @data) 
        while grep $_>0,@data;
1 5 2 6 :  4
0 4 1 5 :  4
-1 3 0 4 :  4
-2 2 -1 3 :  4
-3 1 -2 2 :  4
-4 0 -3 1 :  4

my @data = qw[1 2 5 6]; 
print "\n@data : ", overlap @data 
    and @data = map($_-1, @data) 
        while grep $_>0,@data;
1 2 5 6 :  -2
0 1 4 5 :  -2
-1 0 3 4 :  -2
-2 -1 2 3 :  -2
-3 -2 1 2 :  -2
-4 -3 0 1 :  -2

my @data = qw[2 5 1 6]; 
print "\n@data : ", overlap @data 
    and @data = map($_-1, @data) 
        while grep $_>0,@data;
2 5 1 6 :  4
1 4 0 5 :  4
0 3 -1 4 :  4
-1 2 -2 3 :  4
-2 1 -3 2 :  4
-3 0 -4 1 :  4

my @data = qw[1 6 2 5]; 
print "\n@data : ", overlap @data 
    and @data = map($_-1, @data) 
        while grep $_>0,@data;
1 6 2 5 :  4
0 5 1 4 :  4
-1 4 0 3 :  4
-2 3 -1 2 :  4
-3 2 -2 1 :  4
-4 1 -3 0 :  4
[download]

Which I thought (wrongly) meant I'd covered all the edge cases. A bad testcse is worse than no testcase.

Here's a fixed but it's not nearly so pretty, nor efficient. It's dubious if this is any better than your original.

#! perl -slw
use strict;

sub overlap{
    @_ = map{abs $_ } @_ if 4 == grep $_ < 0, @_;
    local $_ = ($_[1] < $_[3] ? $_[1]: $_[3]) - ($_[0] > $_[2] ? $_[0]
+ : $_[2]) ;
    $_ > -1 ? $_ : 0;
}

while(<DATA>) {
    chomp;
    print,next unless length;
    print $_, ' = ', overlap( ( split(' ') )[0 .. 3] );
}

print'';

my @data = qw[1 5 2 6];

print "@data : ", overlap @data
    and @data = map($_-1, @data)
        while grep $_>0,@data;

__DATA__
1 2 2 3 (0)
2 3 1 2 (0)
1 2 3 4 (0)
1 3 2 4 (1)
1 4 2 3 (1)
2 4 1 3 (1)
3 4 1 2 (0)

-2 -1 1 2 (0)
-2 1 -1 2 (2)
-2 2 -1 1 (2)
-1 2 -2 1 (2)
1 2 -2 -1 (0)

-2 -3 -1 -2 (0)
-1 -2 -3 -4 (0)
-1 -3 -2 -4 (1)
-1 -4 -2 -3 (1)
-2 -4 -1 -3 (1)
-3 -4 -1 -2 (0)
[download]

C:\test>232325
1 2 2 3 (0) = 0
2 3 1 2 (0) = 0
1 2 3 4 (0) = 0
1 3 2 4 (1) = 1
1 4 2 3 (1) = 1
2 4 1 3 (1) = 1
3 4 1 2 (0) = 0

-2 -1 1 2 (0) = 0
-2 1 -1 2 (2) = 2
-2 2 -1 1 (2) = 2
-1 2 -2 1 (2) = 2
1 2 -2 -1 (0) = 0

-2 -3 -1 -2 (0) = 0
-1 -2 -3 -4 (0) = 0
-1 -3 -2 -4 (1) = 1
-1 -4 -2 -3 (1) = 1
-2 -4 -1 -3 (1) = 1
-3 -4 -1 -2 (0) = 0

1 5 2 6 : 3
0 4 1 5 : 3
-1 3 0 4 : 3
-2 2 -1 3 : 3
-3 1 -2 2 : 3
-4 0 -3 1 : 3

C:\test>
[download]

I think I tried all the variations this time, but as always, better minds and different eyes may see it differently.