On Fri, May 17, 2019 at 5:44 PM Victor Khomenko <victor.khomenko@newcastle.ac.uk> wrote:
While you are at this, maybe you could also implement SVA's ##[*] and ##[+] (as sugar for ##[0:$] and ##[1:$]) for completeness?
Yes I can. I have seen those operators on a couple of presentations online, but could not find any mention of them in the language reference manual.
Please also note that the equality a ##[i:j] b = a:([*i:j];b) is wrong (despite being recommended in some textbooks!), e.g.
Ouch. So currently, assuming j>0, I use these rules a ##[i:j] b = a;[*i-1:j-1];b if i > 0 a ##[0:j] b = a:([*0:j];b) // wrong if a accepts [*0] Do you agree with the following replacement? a ##[i:j] b = a;[*i-1:j-1];b if i > 0 a ##[0:j] b = a:([*0:j];b) if a rejects [*0] a ##[0:j] b = (a;[*0:j]):b if a accepts [*0] but b rejects [*0] a ##[0:j] b = (a:b) | (a;[*0:j-1];b) if a and b both reject [*0] The goal is just to avoid duplicating a and b when possible. -- Alexandre Duret-Lutz