On Dec 14, 2006, at 11:20 PM, Matisse Enzer wrote:
On Dec 14, 2006, at 3:05 PM, Michael G Schwern wrote:
Matisse Enzer wrote:
sub complexity_of_six {
my $bar = shift;
my $total = 0;
my $type = ref $bar;
if ( ! $type ) {
$total = $bar;
}
elsif ( $type eq 'ARRAY' ) {
foreach my $baz ( @{$bar} ) {
$total += $baz;
}
}
else {
confess("Don't know how to handle refs to $type");
}
return $total;
}
I'm missing something, I only count 3 paths. if, elsif and else.
The confess() might count (die/throw exception) but its the only
path through the else. Same with the loop, its the only path
through the elsif.
I over-simplified my explanation. It's not exactly paths-through-
the-code.
The complexity "points" in that example are:
1 for the subroutine itself.
1 for the "if"
1 for the ! logic operator in the if condition
1 for the elsif
1 for the "ne" logic operator in the elsif condition
1 for the else
----
6 total
That can't be right. Negation does not contribute to complexity.
Instead, I believe it is the for loop and the exit points that are
increasing your count. Consider rewriting the for as ifs and gotos:
sub complexity_of_six {
my $bar = shift;
my $total = 0;
my $type = ref $bar;
if ( ! $type ) {
$total = $bar;
}
elsif ( $type eq 'ARRAY' ) {
my $_i = 0;
LOOP:
goto DONE if ($_i >= @{$bar});
my $baz = $bar->[$_i];
$total += $baz;
$_i++;
goto LOOP;
DONE:
}
else {
confess("Don't know how to handle refs to $type");
}
return $total;
}
Then the decision points are:
1) if ( ! $type )
2) if ( $type eq 'ARRAY')
3) if ($_i >= @{$bar})
4) else
and the end points are
1) confess("Don't know how to handle refs to $type");
2) return $total
So I actually count a complexity of 7 (num decision points + num
endpoints + 1) if I've understood the wiki definition correctly.
http://en.wikipedia.org/wiki/Cyclomatic_complexity
Chris
--
Chris Dolan, Software Developer, Clotho Advanced Media Inc.
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