Hi All, My last question got me thinking (always dangerous): an expression that doubles a list and takes the 2nd element (index 1), I assume, goes through the following steps. double (double [1,2,3,4]) !! 1 double ( 1*2 : double [2,3,4]) !! 1 1*2*2 : double ( double [2,3,4] ) !! 1 1*2*2 : double ( 2*2 : double [3,4] ) !! 1 1*2*2 : 2*2*2 : double ( double [3,4] ) !! 1 2*2*2 8 Up until the element requested all the proceeding elements have to have the expression formed as Haskell process the calculation. Now, I want to compare this to using function composition ( double . double ) [ 1 ,2, 3, 4 ] !! 1 This is the bit I'm unsure of - what does the composition look like. It is easy to see that it could be simplified to something like: ( double . double) (x:xs) = x*4 : (double . double) xs This would mean the steps could be written (double . double) [ 1,2,3,4 ] !! 1 (1*4 : (double.double) [2,3,4]) !! 1 (1*4 : 2*4 : (double.double) [ 3,4 ]) !! 1 2*4 8 Ignoring the start and end steps, it will be half the number of steps compared to the application version. Significant then, over long lists. So is this true, are composite functions simplified in Haskell in general so that they may have improved performance over function application? I've seen posts on the internet that say it's a matter of style only: http://stackoverflow.com/questions/3030675/haskell-function-composition-and-.... But my reasoning suggests it could be more than that. Perhaps, function application is similarly optimised - maybe by replacing all functions applications with composition and then simplifying? Or maybe the simplifying/optimisation step never happens? As you can see I'm just guessing at things :-) But it's nice to wonder. Many thanks for any pointers, Matt.