Hi, I have been playing around with Haskell for about a month now and reading the nice book "Real World Haskell." My main reason for learning Haskell is that I want to code up some machine learning projects (heavy use of matricies). Normally, I use Mathematica and mix it with C++, but Mathematica is proprietary, slow, and can't produce executable while C++ is verbose. Learning Haskell has been fun; however, I have been a little worried that I will sacrifice too much performance when coding in Haskell. So, I recently asked one of my friends the following question, "Say you had a C program. Can you always convert it to Haskell in such a way that the compiled Haskell is not too slow and does not need too much memory?" Supposing that too slow means slower than 1/4 the speed of C and too much memory means twice the memory of C. Hopefully, someone that knows Haskell well can comment on this question. I am not sure, but I think the answer is yes, such a conversion can always be done and creating a C to Haskell compiler with the above performance constraints is not extremely hard. I started thinking about how a compiler might convert a simple C program into Haskell. Below I will paste a C program and the compile-by-hand Haskell code. It seems to me that the ideas I used to create the Haskell code can be implemented in a compiler that converts a simple subset of C into Haskell. I was thinking about restricting the C to one data type 32bit-integers, arithmetic (+-*/%), assignment (=), comparison (<,>,==,<=,>=), the if-condition-codeblock construct, and the while-condition-codeblock construct. (I would also like to do integer arrays, but I have not read about mutable arrays and monads yet.) Any comments? Cheers, Hein H. --------------------C to Haskell------------------------ module Main where -- Convert a C-Program line by line into Haskell -- -- 1-- #include <stdio.h> -- 2-- void main() -- 3-- { -- 4-- long i,j; -- 5-- i=7; -- 6-- j=0; -- 7-- while(j<10000) -- 8-- { -- 9-- if ((i % 17)== 11) --10-- i = i*2; --11-- if ((i % 35)== 12) --12-- i = i+13; --13-- if (i> 1000) --14-- i = i - 1000; --15-- i++; --16-- j++; --17-- } --18-- printf("%ld", i); --19--} --assignment statements line5 (i,j) = (7,j) line6 (i,j) = (i,0) line10 (i,j) = (i*2,j) line12 (i,j) = (i+13,j) line14 (i,j) = (i-1000,j) line15 (i,j) = (i+1,j) line16 (i,j) = (i,j+1) --while statement line7test (i,j) = j >=100000000 line7 (i,j) = until line7test body9To16 (i,j) --if statements line9 (i,j) = if ((mod i 17) == 11) then line10 (i,j) else (i,j) line11 (i,j) = if ((mod i 35) == 12) then line12 (i,j) else (i,j) line13 (i,j) = if (i>1000) then line14 (i,j) else (i,j) -- code blocks mainprogram :: (Int, Int) -> (Int, Int) mainprogram = line7 . line6 . line5 body9To16 = line16 . line15 . line13 . line11 . line9 main :: IO () main = putStrLn(show(mainprogram (0,0))) -----------------end of program-------------------------