As has been explained quite explicitly twice already, you call it as a sub by using the full explicit name of the subroutine:
$z = infix:<+^>($x, $y) On Sun, 19 Jan 2020 at 07:22, ToddAndMargo via perl6-users < perl6-us...@perl.org> wrote: > On 2020-01-18 06:09, Tobias Boege wrote: > > On Fri, 17 Jan 2020, ToddAndMargo via perl6-users wrote: > >> Hi All, > >> > >> https://docs.raku.org/routine/+$CIRCUMFLEX_ACCENT > >> > >> (Operators) infix +^ยง > >> > >> multi sub infix:<+^>($a, $b --> Int:D) > >> > >> Integer bitwise XOR operator: Coerces both arguments to Int and does a > >> bitwise XOR (exclusive OR) operation. > >> > >> > >> $ p6 'my uint8 $x=0b0101_0111; my uint8 $y = 0b0000_1111; my uint8 $z = > >> +^($x, $y ); say "0", $x.base(2); say "0000", $y.base(2); say > $z.base(2);' > >> > >> 01010111 > >> 00001111 > >> 11111101 > >> > >> > >> XOR > >> A B A xor B > >> 0 0 0 > >> 1 0 1 > >> 0 1 1 > >> 1 1 0 > >> > >> That s not what I am seeing above. > >> > >> What am I doing wrong this time? > >> > >> And who set the high bit making $z negative? > >> > > > > As was already mentioned, if you want to use the &infix:<+^> operator, > > you have to either call it by its full name as a sub: > > > > &infix:<+^>($x, $y) > > > > or you have to use it according to its syntax category, as an infix: > > > > $x +^ $y > > > > When you write +^($x,$y), its cousin &prefix:<+^> is called instead, > > because now you use +^ as a prefix operator. That one performs bitwise > > negation on its argument coerced to Int, and since you pass it a list > > of length two, you actually evaluate +^2. And that's how you get the > > high bit and a value of -3. > > > > Regards, > > Tobias > > > > Hi Tobias, > > This works perfectly using the syntax $a = $b +^ $c > > p6 'my uint8 $x=0b0101_0111; my uint8 $y = 0b0000_1111; my uint8 $z = $x > +^ $y; say "0", $x.base(2); say "0000", $y.base(2); say "0",$z.base(2);' > > 01010111 # $x > 00001111 # $y > 01011000 # z = x xor y > > > But what is now confusing me is $c = +^( $a, $b ) > > p6 'my uint8 $x=0b0101_0111; my uint8 $y = 0b0000_1111; my uint8 $z = > +^($x, $y); say "0", $x.base(2); say "0000", $y.base(2); say $z.base(2);' > > 01010111 # $x > 00001111 # $y > 11111101 # z is not x xor y > > Which is clearly not correct. The manual states: > > Integer bitwise XOR operator: Coerces both > arguments to Int and does a bitwise XOR > (exclusive OR) operation. > > In my example run line, I purposefully left the > high bit off to keep overly helpful Raku from thinking > I had a negative number when it annoyingly overrules > my choice of native type. (Ordinarily I like this > feature, except when doing bitwise operations, I > wish I could turn it off.) > > So I am confused. > > -T >
