On 10/3/18 7:16 PM, ToddAndMargo wrote:
Bitwise "IN" (Does y exist in x):$ p6 'my $x=0b1001; my $y=0b0101; my $z=$x +& $y; say so $y == $z;' False $ p6 'my $x=0b1001; my $y=0b1001; my $z=$x +& $y; say so $y == $z;' True p5 not figured out yet
$ p5 'my $x = 0b1001; my $y = 0b1000; say qw(false true)[($x & $y) == $y]'
true$ p5 'my $x = 0b1001; my $y = 0b0100; say qw(false true)[($x & $y) == $y]'
false