On Mon, Jan 26, 2015, at 08:43 PM, Animesh Saxena wrote:

I stumbled across this example as a shiny way of explaining why Lazy eval matters....

 

    fact = fix $ \f n -> if n == 0 then 1 else n * f (n-1)

fix f = f (fix f)

 

With lazy eval, I get 

 

fact 4 = fix $ \f 4 (......)

 

Not quite. Let's go one step at a time:

 

fact 4 = (fix $ \f n -> if n == 0 then 1 else n * f (n - 1)) 4

 

Now we can bring in the definition of fix, substituting the argument to fix for all of the occurrences of f in the definition of fix. Notice that we can't substitute the 4 for n yet.

 

fact 4 = ((\f n -> if n == 0 then 1 else n * f (n - 1)) (fix $ \f n -> if n == 0 then 1 else n * f (n - 1))) 4

 

I think if you are very patient and methodical about performing the substitutions (and writing them out fully - no dot-dot-dot), then you'll figure out how it all works.

 

-Karl

 

_______________________________________________
Beginners mailing list
Beginners@haskell.org
http://www.haskell.org/mailman/listinfo/beginners