Hi Mateusz,
IORef and STRef are boxed references. That is, they are a mutable cell
that contains a pointer to some immutable Haskell value. When you
increment a (STRef Int), you first dereference the pointer, allocate a
new immutable heap object to represent the new integer value, then
mutate the reference to point to the new object. This costs much more
than updating a plain mutable integer.
As far as I know there is no particularly popular library that
provides mutable references like this. As a workaround, you can create
a 1-sized unboxed mutable vector using the vector package, and use it
like a reference.
On 3 December 2015 at 01:10, Mateusz Kłoczko
Hello!
I've performed a few simple tests using Haskell and C++ on primitives. I've compilled all Haskell programs with -O2 optimizations, and C++ ones with -O3 ghc version - 7.10.2, gcc version : 5.1.1
I'm sending the codes in the zip file.
Problem1 - 100 000 000 iterations.
Time of execution (in seconds): Int pure tail recursion: 6.911011299962411e-2 Int# pure tail recursion : 4.587398100011342e-2 IORef for loop 1.1533970820000832 IORef 'tail' recursion 1.0696569040001123 STRef for loop 1.1545546840006864 STRef tail recursion 1.1110019479992843 C++ : 2.7e-07
On this one, g++ manages to eliminate the loop entirely, but GHC doesn't.
The llvm version could be as fast as C++ one in this problem.
Buuut... then there's problem 2 (using if or case) - 100 000 000 iterations: Int# tail recursion 1.315346227000191 IORef for loop: 2.6442542390004746 STRef for loop: 2.669217500999366 C++: 0.158056
Here haskell is about 9 times slower than C++.
The main difference on this comes from the fact that GHC does not optimize (n `remInt#` 2#) into (n `andI#` 1#). There is a ticket [0] that contains some discussion of this issue. [0]: https://ghc.haskell.org/trac/ghc/ticket/5615 Hope this helps, Takano Akio