SNA=Shah Namrata Abhaykumar
SNA Consider the following eg.
SNA
SNA Using the above tanslation ,
SNA
SNA case (1,2) of { ~(a,b) = a + b ; _ -> 0 }
SNA
SNA == let { y = (1,2) }
SNA in
SNA let { x1' = case y of { (a,b) -> a }}
SNA in
SNA let { x2' = case y of { (a,b) -> b }}
SNA in
SNA x1' + x2'
SNA
SNA So, when x1' + x2' is evaluated,
SNA Is (1,2) pattern matched against (a,b) twice -- once for x1' and once
SNA for x2' ??
It is matched twice, but this doesn't matter -- matching is side-effect-free.
SNA But,
SNA When (1,2) is pattern matched with (a,b) for the first time ,
SNA bindings for both a and b can be created. So when x2'is needed,
SNA pattern matching need not be done - the binding created
SNA during the first pattern match can be used.
In your example, yes, but consider the following:
case (1,2,undef) of { ~(a,b,~(c,d)) = a+b; _ -> 0 }
You have to delay the binding of c and d,
because the subpattern is still irrefutable after the translation
(and not a variable).
---
Stefan Kahrs
