On Thu, Feb 10, 2005 at 10:42:34AM +0000, Thomas Yandell wrote:
> Is the following comment correct?
>
> my $x = any(2,3,4,5) and any(4,5,6,7); # $x now contains any(4,5)
Short answer: I don't think so.
Long answer: I tend to get very lost when dealing with junctions, so
I can be completely wrong. However, watch the precedence and meanings
of the operators here -- I would think that
my $x = any(2,3,4,5) and any(4,5,6,7);
results in $x containing any(2,3,4,5), just as
my $y = 2 and 3;
results in $y containing 2 (since C<and> has lower precedence than C<=>).
Even if you fixed the =/and precedence with parens, to read
my $x = (any(2,3,4,5) and any(4,5,6,7));
then I think the result is still that $x contains any(4,5,6,7).
It gets interpreted as (from S09):
$x = any( 2 and 4, # 4
2 and 5, # 5
2 and 6, # 6
2 and 7, # 7
3 and 4, # 4
3 and 5, # 5
# etc...
5 and 6, # 6
5 and 7, # 7
);
which ultimately boils down to any(4,5,6,7).
Pm
