21 Feb
2015
21 Feb
'15
3:58 a.m.
Me too. Last I heard, there wasn't any code that depended on foldl being lazy, and it doesn't really seem possible. http://www.well-typed.com/blog/90/ On Fri, Feb 20, 2015 at 1:13 AM, Roman Cheplyaka <roma@ro-che.info> wrote:
I'd be curious to see a (non-contrived) example.
On 20/02/15 09:05, David Feuer wrote:
Probably not. There's real code that depends on the current foldl semantics.
On Wed, Feb 18, 2015 at 10:40 AM, Joe Hillenbrand <joehillen@gmail.com> wrote:
Is foldl = foldl' ever going to happen?