> On 27 May 2017, at 23:22, Elizabeth Mattijsen <l...@dijkmat.nl> wrote: >> On 27 May 2017, at 19:33, Marcel Timmerman <mt1...@gmail.com> wrote: >> In perl6 version 2017.04.3-287-g3e7675a built on MoarVM version >> 2017.04-64-g6d5ea04 >> implementing Perl 6.c. I observe the following; >> >> my Num $num = Inf; >> my FatRat $f = $num.FatRat; >> Type check failed in assignment to $f; expected FatRat but got >> Rational[Num,Int] (?) >> in block <unit> at <unknown file> line 1 >> >> Inf.WHAT; # (Num) >> Inf.FatRat.WHAT # (Rational[Num,Int]) >> >> It boils down to not converting to FatRat while a defined number does; >> my Num $num = 2.3e3; >> $num.FatRat.WHAT; # (FatRat) >> >> >> This is the same for NaN. >> >> If this is defined behavior, then there is no way to set NaN or Inf for >> FatRat (and for Rat too). Is there another way to accomplish this? > > This feels like a bug. Could you rakudobug it so it won’t fall through the > cracks? > > FWIW, the golf is: > > $ 6 'dd FatRat.new(Inf)' > Type check failed in binding to parameter 'nu'; expected Int but got Num (Inf) > in block <unit> at -e line 1
Please disregard the above. This is a problem that basically comes down to: $ 6 'say Int.Range.max.WHAT' (Num) In other words: the maximum value of an Int is *not* an Int. To be able to support Inf in rationals, we use a hack to create a special type of Rat that allows for a Num numerator. https://github.com/rakudo/rakudo/blob/nom/src/core/Num.pm#L44 I’m not seeing a way around this at the moment. But then I’m really tired now :-)
