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

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