22 Mar
2005
22 Mar
'05
12:10 p.m.
Hello again, Sorry, I made a little mistake.
a :: Int a = let ?foo = 0 in b where b :: (?foo :: Int) => Int b = let ?foo = 1 in c where c = ?foo The meaning of this code depends on the flag -f(no)-monomorphism-restriction since with the monomorphism turned on, `c' gets the monomorphic type `Int', and the `?foo' in the definition of `c' refers to the implicit parameter of `b', so `a' evaluates to `0'. On the other hand, without the monomorphism restriction, the type of `c' becomes `(?foo :: Int) => Int', and it is easy to see that `a' evaluates to `0'. In this case, `a' of course evaluates to `1'.
Thomas