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?
>>