On 3/20/06, Sebastian Sylvan <[EMAIL PROTECTED]> wrote: > On 3/19/06, Manuel M T Chakravarty <[EMAIL PROTECTED]> wrote: > > Loosely related to Ticket #76 (Bang Patterns) is the question of whether > > we want the language to include strict tuples. It is related to bang > > patterns, because its sole motivation is to simplify enforcing > > strictness for some computations. Its about empowering the programmer > > to choose between laziness and strictness where they deem that necessary > > without forcing them to completely re-arrange sub-expressions (as seq > > does). > > > > So what are strict tupples? If a lazy pair is defined in pseudo code as > > > > data (a, b) = (a, b) > > > > a strict pair would be defined as > > > > data (!a, b!) = ( !a, !b ) > > > > Ie, a strict tuple is enclosed by bang parenthesis (! ... !). The use > > of the ! on the rhs are just the already standard strict data type > > fields. > > > > Maybe I've missed something here. But is there really any reasonable > usage cases for something like: > > f !(a,b) = a + b > > in the current bang patterns proposal? > > I mean, would anyone really ever want an explicitly strict (i.e. using > extra syntax) tuple with lazy elements? > > Couldn't the syntax for strict tuples be just what I wrote above > (instead of adding weird-looking exclamation parenthesis). > > I'm pretty sure that most programmers who would write "f !(a,b) = ..." > would expect the tuple's elements to be forced (they wouldn't expect > it to do nothing, at least).. In fact !(x:xs) should mean (intuitively > to me, at least) "force x, and xs", meaning that the element x is > forced, and the list xs is forced (but not the elements of the xs). > > Couldn't this be generalised? A pattern match on any constructor with > a bang in front of it will force all the parts of the constructor > (with seq)? > > So: > f !xs = b -- gives f xs = xs `seq` b, like the current proposal > f !(x:xs) = b -- gives f (x:xs) = x `seq` xs `seq` b, unlike the > current proposal? > > The latter would then be equal to > > f (!x:xs) = b
I mean f (!x:!xs) = b /S -- Sebastian Sylvan +46(0)736-818655 UIN: 44640862 _______________________________________________ Haskell-prime mailing list Haskell-prime@haskell.org http://haskell.org/mailman/listinfo/haskell-prime