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