High-Performance Haskell

High-Performance Haskell

Johan Tibell
johan.tibell@gmail.com

2010-10-01

Welcome!

A few things about this tutorial:
Stop me and ask questions—early and often
I assume no prior Haskell exposure

Sequential performance is still important

Parallelism is not a magic bullet:
The speedup of a program using multiple processors is limited
by the time needed for the sequential fraction of the program.
(Amdahl’s law)
We want to make eﬃcient use of every core.

Caveats

The usual caveats about performance optimizations:
Improvements to the compiler might make some optimizations
redundant. Write benchmarks to detect these cases.
Some optimizations are compiler (i.e. GHC) speciﬁc
That being said, many of these optimizations have remained valid
over a number of GHC releases.

Software prerequisites

The Haskell Platform:
Download installer for Windows, OS X, or Linux here:
http://hackage.haskell.org/platform
The Criterion benchmarking library:

cabal install -f-Chart criterion

Outline

Introduction to Haskell
Lazy evaluation
Reasoning about space usage
Benchmarking
Making sense of compiler output
Proﬁling

Haskell in 10 minutes

Our ﬁrst Haskell program sums a list of integers:
sum : : [ I n t ] −> I n t
sum [ ] = 0
sum ( x : x s ) = x + sum x s

main : : IO ( )
main = p r i n t (sum [ 1 . . 1 0 0 0 0 ] )

Type signatures

Deﬁnition
A type signature describes the type of a Haskell expression:

sum : : [ I n t ] −> I n t

Int is an integer.
[a] is a list of as
So [ Int ] is a list of integers
−> denotes a function.
So sum is a function from a list of integers to an integer.

Defining a function

Functions are defined as a series of equations, using pattern
matching:
sum [ ] = 0
The list is defined recursively as either
an empty list, written as [] , or
an element x, followed by a list xs.
[] is pronounced “nil” and : is pronounced “cons”.

Function application

Function application is indicated by juxtaposition:
main = p r i n t (sum [ 1 . . 1 0 0 0 0 ] )

[1..10000] creates a list of 10,000 integers from 1 to 10,000.
We apply the sum function to the list and then apply the
result to the print function.
We say that we apply rather then call a function:
Haskell is a lazy language
The result may not be computed immediately

Compiling and running our program

Save the program in a ﬁle called Sum.hs and then compile it using
ghc:

$ ghc -O --make Sum.hs
[1 of 1] Compiling Main ( Sum.hs, Sum.o )
Linking Sum ...

Now lets run the program

$ ./Sum
50005000

Defining our own data types

Data types have one or more constructors, each with zero or more
arguments (or fields).
data Shape = C i r c l e Double
| R e c t a n g l e Double Double
And a function over our data type, again defined using pattern
matching:
a r e a : : Shape −> Double
area ( C i r c l e r ) = r ∗ r ∗ 3.14
area ( Rectangle w h) = w ∗ h
Constructing a value uses the same syntax as pattern matching:
area ( Rectangle 3.0 5.0)

Back to our sum function

Our sum has a problem. If we increase the size of the input
main = p r i n t (sum [ 1 . . 1 0 0 0 0 0 0 0 ] )
and run the program again

Linking Sum ...
$ ./Sum
Stack space overflow: current size 8388608 bytes.
Use ‘+RTS -Ksize -RTS’ to increase it.

Tail recursion

Our function creates a stack frame for each recursive call,
eventually reaching the predeﬁned stack limit.
Must do so as we still need to apply + to the result of the call.
Make sure that the recursive application is the last thing in the
function
sum : : [ I n t ] −> I n t
sum x s = sum ’ 0 x s
where
sum ’ a c c [ ] = acc
sum ’ a c c ( x : x s ) = sum ’ ( a c c + x ) x s

Polymorphic functions
Many functions follow the same pattern. For example,
product : : [ I n t ] −> I n t
product x s = product ’ 1 x s
where
product ’ a c c [ ] = acc
product ’ a c c ( x : x s ) = product ’ ( a c c ∗ x ) x s
is just like sum except we replace 0 with 1 and + with *. We can
generalize sum and product to
f o l d l : : ( a −> b −> a ) −> a −> [ b ] −> a
foldl f z [] = z
f o l d l f z ( x : xs ) = f o l d l f ( f z x ) xs

sum = f o l d l (+) 0
product = f o l d l ( ∗ ) 1

Summing some numbers...

Using our new deﬁnition of sum, lets sum all number from 1 to
1000000:

Linking Sum ...
$ ./Sum

What went wrong this time?

Laziness

Haskell is a lazy language
Functions and data constructors don’t evaluate their
arguments until they need them
cond : : Bool −> a −> a −> a
cond True t e = t
cond F a l s e t e = e

Same with local deﬁnitions
abs : : I n t −> I n t
abs x | x > 0 = x
| otherwise = neg x
where n e g x = negate x

Why laziness is important

Laziness supports modular programming
Programmer-written functions instead of built-in language
constructs
( | | ) : : Bool −> Bool −> Bool
True | | = True
False | | x = x

Laziness and modularity

Laziness lets us separate producers and consumers and still get
eﬃcient execution:
Generate all solutions (a huge tree structure)
Find the solution(s) you want

nextMove : : Board −> Move
nextMove b = s e l e c t M o v e a l l M o v e s
where
allMoves = allMovesFrom b
The solutions are generated as they are consumed.

Back to our missbehaving function

How does evaluation of this expression proceed?
sum [ 1 , 2 , 3 ]
Like this:

sum [1,2,3]
==> foldl (+) 0 [1,2,3]
==> foldl (+) (0+1) [2,3]
==> foldl (+) ((0+1)+2) [3]
==> foldl (+) (((0+1)+2)+3) []
==> ((0+1)+2)+3
==> (1+2)+3
==> 3+3
==> 6

Thunks

A thunk represents an unevaluated expression.
GHC needs to store all the unevaluated + expressions on the
heap, until their value is needed.
Storing and evaluating thunks is costly, and unnecessary if the
expression was going to be evaluated anyway.
foldl allocates n thunks, one for each addition, causing a
stack overﬂow when GHC tries to evaluate the chain of
thunks.

Controlling evaluation order

The seq function allows to control evaluation order.
seq : : a −> b −> b
Informally, when evaluated, the expression seq a b evaluates a and
then returns b.

Weak head normal form

Evaluation stops as soon as a data constructor (or lambda) is
reached:

ghci> seq (1 ‘div‘ 0) 2
*** Exception: divide by zero
ghci> seq ((1 ‘div‘ 0), 3) 2
2

We say that seq evaluates to weak head normal form (WHNF).

Weak head normal form

Forcing the evaluation of an expression using seq only makes sense
if the result of that expression is used later:
l e t x = 1 + 2 i n seq x ( f x )
The expression
p r i n t ( seq ( 1 + 2 ) 3 )
doesn’t make sense as the result of 1+2 is never used.

Exercise

Rewrite the expression
(1 + 2 , ’ a ’ )
so that the component of the pair is evaluated before the pair is
created.

Solution

Rewrite the expression as
l e t x = 1 + 2 i n seq x ( x , ’ a ’ )

A strict left fold

We want to evaluate the expression f z x before evaluating the
recursive call:
f o l d l ’ : : ( a −> b −> a ) −> a −> [ b ] −> a
foldl ’ f z [ ] = z
foldl ’ f z ( x : xs ) = l e t z ’ = f z x
i n seq z ’ ( f o l d l ’ f z ’ x s )

Summing numbers, attempt 2

How does evaluation of this expression proceed?

foldl’ (+) 0 [1,2,3]

Like this:

foldl’ (+) 0 [1,2,3]
==> foldl’ (+) 1 [2,3]
==> foldl’ (+) 3 [3]
==> foldl’ (+) 6 []
==> 6

Sanity check:

ghci> print (foldl’ (+) 0 [1..1000000])
500000500000

Computing the mean

A function that computes the mean of a list of numbers:
mean : : [ Double ] −> Double
mean x s = s / f r o m I n t e g r a l l
where
( s , l ) = f o l d l ’ s t e p (0 , 0) xs
s t e p ( s , l ) a = ( s+a , l +1)
We compute the length of the list and the sum of the numbers in
one pass.

$ ./Mean

Didn’t we just ﬁx that problem?!?

seq and data constructors

Remember:
Data constructors don’t evaluate their arguments when
created
seq only evaluates to the outmost data constructor, but
doesn’t evaluate its arguments
Problem: foldl ’ force the evaluation of the pair constructor, but
not its arguments, causing unevaluated thunks build up inside the
pair:

(0.0 + 1.0 + 2.0 + 3.0, 0 + 1 + 1 + 1)

Forcing evaluation of constructor arguments

We can force GHC to evaluate the constructor arguments before
the constructor is created:
where
( s , l ) = f o l d l ’ s t e p (0 , 0) xs
step (s , l ) a = let s ’ = s + a
l ’ = l + 1
i n seq s ’ ( seq l ’ ( s ’ , l ’ ) )

Bang patterns

Bang patterns is a concise way to express that an argument should
be evaluated.
{−# LANGUAGE B a n g P a t t e r n s #−}

where
( s , l ) = f o l d l ’ s t e p (0 , 0) xs
s t e p ( ! s , ! l ) a = ( s + a , l + 1)
s and l are evaluated before the right-hand side of step is
evaluated.

Strictness

We say that a function is strict in an argument, if evaluating the
function always causes the argument to be evaluated.
n u l l : : [ a ] −> Bool
n u l l [ ] = True
null = False
null is strict in its ﬁrst (and only) argument, as it needs to be
evaluated to pick a return value.

Strictness - Example

cond is strict in the ﬁrst argument, but not in the second and third
argument:
cond : : Bool −> a −> a −> a
cond True t e = t
cond F a l s e t e = e
Reason: Each of the two branches only evaluate one of the two
last arguments to cond.

Strict data types

Haskell lets us say that we always want the arguments of a
constructor to be evaluated:
data P a i r S a b = PS ! a ! b
When a PairS is evaluated, its arguments are evaluated.

Strict pairs as accumulators

We can use a strict pair to simplify our mean function:
where
PS s l = f o l d l ’ s t e p ( PS 0 0 ) x s
s t e p ( PS s l ) a = PS ( s + a ) ( l + 1 )
Tip: Prefer strict data types when laziness is not needed for your
program to work correctly.

Reasoning about laziness

A function application is only evaluated if its result is needed,
therefore:
One of the function’s right-hand sides will be evaluated.
Any expression which value is required to decide which RHS
to evaluate, must be evaluated.
By using this “backward-to-front” analysis we can ﬁgure which
arguments a function is strict in.

Reasoning about laziness: example

max : : I n t −> I n t −> I n t
max x y
| x > y = x
| x < y = y
| otherwise = x −− a r b i t r a r y

To pick one of the three RHS, we must evaluate x > y.
Therefore we must evaluate both x and y.
Therefore max is strict in both x and y.

Poll

data BST = L e a f | Node I n t BST BST

i n s e r t : : I n t −> BST −> BST
insert x Leaf = Node x L e a f L e a f
i n s e r t x ( Node x ’ l r )
| x < x’ = Node x ’ ( i n s e r t x l ) r
| x > x’ = Node x ’ l ( i n s e r t x r )
| o t h e r w i s e = Node x l r
Which arguments is insert strict in?
None
1st
2nd
Both

Solution

Only the second, as inserting into an empty tree can be done
without comparing the value being inserted. For example, this
expression
i n s e r t ( 1 ‘ div ‘ 0 ) L e a f
does not raise a division-by-zero expression but
i n s e r t ( 1 ‘ div ‘ 0 ) ( Node 2 L e a f L e a f )
does.

Some other things worth pointing out

insert x l is not evaluated before the Node is created, so it’s
stored as a thunk.
Most tree based data structures use strict sub-trees:
data S e t a = Tip
| Bin ! S i z e a ! ( S e t a ) ! ( S e t a )

Reasoning about space usage

Knowing how GHC represents values in memory is useful because
it allows us to approximate memory usage, and
it allows us to count the number of indirections, which aﬀect
cache behavior.

Memory layout

Here’s how GHC represents the list [1,2] in memory:

(:) (:) []

I# 1 I# 2

Each box represents one machine word
Arrows represent pointers
Each constructor has one word overhead for e.g. GC
information

Memory usage for data constructors

Rule of thumb: a constructor uses one word for a header, and one
word for each ﬁeld. So e.g.
data Uno = Uno a
data Due = Due a b
an Uno takes 2 words, and a Due takes 3.
Exception: a constructor with no ﬁelds (like Nothing or
True) takes no space, as it’s shared among all uses.

Unboxed types

GHC deﬁnes a number of unboxed types. These typically represent
primitive machine types.
By convention, the names of these types end with a #.
Most unboxed types take one word (except e.g. Double# on
32-bit machines)
Values of unboxed types are never lazy.
The basic types are deﬁned in terms unboxed types e.g.
data I n t = I# I n t#

We call types such as Int boxed types

Poll

How many machine words is needed to store a value of this data
type:
data I n t P a i r = IP I n t I n t

3?
5?
7?
9?

IntPair memory layout

IP I# Int#

I# Int#

So an IntPair value takes 7 words.

Unpacking

GHC gives us some control over data representation via the
UNPACK pragma.
The pragma unpacks the contents of a constructor into the
ﬁeld of another constructor, removing one level of indirection
and one constructor header.
Any strict, monomorphic, single-constructor ﬁeld can be
unpacked.
The pragma is added just before the bang pattern:
data Foo = Foo {−# UNPACK #−} ! SomeType

Unpacking example

data I n t P a i r = IP I n t I n t

IP I# Int#

I# Int#

data I n t P a i r = IP {−# UNPACK #−} ! I n t
{−# UNPACK #−} ! I n t

IP Int# Int#

Benefits of unpacking

When the pragma applies, it offers the following benefits:
Reduced memory usage (4 words in the case of IntPair )
Fewer indirections
Caveat: There are cases where unpacking hurts performance e.g. if
the fields are passed to a non-strict function.

Benchmarking

In principle, measuring code-execution time is trivial:
1. Record the start time.
2. Execute the code.
3. Record the stop time.
4. Compute the time diﬀerence.

A naive benchmark

import time

def bench(f):
start = time.time()
f()
end = time.time()
print (end - start)

Benchmarking gotchas

Potential problems with this approach:
The clock resolution might be too low.
The measurement overhead might skew the results.
The compiler might detect that the result of the function isn’t
used and remove the call completely!
Another process might get scheduled while the benchmark is
running.
GC costs might not be completely accounted for.
Caches might be warm/cold.

Statistically robust benchmarking

The Criterion benchmarking library:
Figures out how many times to run your function
Adjusts for measurement overhead
Computes conﬁdence intervals
Detects outliers
Graphs the results

cabal install -f-Chart Criterion

Benchmarking our favorite function

import C r i t e r i o n . Main

f i b : : I n t −> I n t
fib 0 = 0
fib 1 = 1
f i b n = f i b ( n−1) + f i b ( n−2)

main = d e f a u l t M a i n [
bench ” f i b 10 ” ( whnf f i b 1 0 )
]

Benchmark output, part 1

$ ./Fibber
warming up
estimating clock resolution...
mean is 8.638120 us (80001 iterations)
found 1375 outliers among 79999 samples (1.7%)
1283 (1.6%) high severe
estimating cost of a clock call...
mean is 152.6399 ns (63 iterations)
3 (4.8%) high mild

Benchmark output, part 2

benchmarking fib 10
collecting 100 samples, 9475 iterations each, in
estimated 863.8696 ms
bootstrapping with 100000 resamples
mean: 925.4310 ns, lb 922.1965 ns, ub 930.9341 ns,
ci 0.950
std dev: 21.06324 ns, lb 14.54610 ns, ub 35.05525 ns,
ci 0.950
7 (7.0%) high severe
variance introduced by outliers: 0.997%
variance is unaffected by outliers

Evaluation depth

bench ” f i b 10 ” ( whnf f i b 1 0 )

whnf evaluates the result to weak head normal form (i.e. to
the outmost data constructor).
If the benchmark generates a large data structure (e.g. a list),
you can use nf instead to force the generation of the whole
data structure.
It’s important to think about what should get evaluated in the
benchmark to ensure your benchmark reﬂects the use case you
care about.

Benchmark: creating a list of 10k elements


bench ” whnf ” ( whnf ( r e p l i c a t e n ) ’ a ’ )
, bench ” n f ” ( n f ( r e p l i c a t e n ) ’ a ’ )
]
where n = 10000
The expression replicate n x creates a list of n copies of x.

The diﬀerence between whnf and nf

$ ./BuildList
benchmarking whnf
mean: 15.04583 ns, lb 14.97536 ns, ub 15.28949 ns, ...
std dev: 598.2378 ps, lb 191.2617 ps, ub 1.357806 ns, ...

benchmarking nf
mean: 158.3137 us, lb 158.1352 us, ub 158.5245 us, ...
std dev: 993.9415 ns, lb 834.4037 ns, ub 1.242261 us, ...

Since replicate generates the list lazily, the ﬁrst benchmark only
creates a single element

A note about the whnf function

The whnf function is somewhat peculiar:
It takes the function to benchmark, applied to its ﬁrst n-1
arguments, as its ﬁrst argument
It takes the last, nth, argument as its second argument
For example:
bench ” p r o d u c t ” ( whnf ( f o l d l ( ∗ ) 1 ) [ 1 . . 1 0 0 0 0 ] )
By separating the last arguments from the rest, Criterion tries to
prevent GHC from evaluating the function being benchmarked only
once.

Exercise: foldl vs foldl’
Benchmark the performance of foldl (+) 0 [1..10000] and
foldl ’ (+) 0 [1..10000] . Start with:
import Prelude h i d i n g ( f o l d l )

f o l d l , f o l d l ’ : : ( a −> b −> a ) −> a −> [ b ] −> a
foldl f z = . . .
foldl ’ f z = . . .

bench ” f o l d l ” . . .
, bench ” f o l d l ’ ” . . .
]
Compiling and running:

$ ghc -O --make Fold.hs && ./Fold

Solution


foldl f z [] = z
f o l d l f z ( x : xs ) = f o l d l f ( f z x ) xs

foldl ’ f z [ ] = z
foldl ’ f z ( x : xs ) = l e t z ’ = f z x
i n seq z ’ ( f o l d l ’ f z ’ x s )

bench ” f o l d l ” ( whnf ( f o l d l (+) 0 ) [ 1 . . 1 0 0 0 0 ] )
, bench ” f o l d l ’ ” ( whnf ( f o l d l ’ (+) 0 ) [ 1 . . 1 0 0 0 0 ] )
]

Solution timings

benchmarking foldl
mean: 493.6532 us, lb 488.0841 us, ub 500.2349 us, ...
std dev: 31.11368 us, lb 26.20585 us, ub 42.98257 us, ...

benchmarking foldl’
mean: 121.3693 us, lb 120.8598 us, ub 122.6117 us, ...
std dev: 3.816444 us, lb 1.889005 us, ub 7.650491 us, ...

GHC Core

GHC uses an intermediate language, called “Core,” as its
internal representation during several compilation stages
Core resembles a subset of Haskell
The compiler performs many of its optimizations by
repeatedly rewriting the Core code

Why knowing how to read Core is important

Reading the generated Core lets you answer many questions, for
example:
When are expressions evaluated
Is this function argument accessed via an indirection
Did my function get inlined

Convincing GHC to show us the Core

Given this “program”
module Sum where

import Prelude h i d i n g (sum)

sum : : [ I n t ] −> I n t
sum [ ] = 0
we can get GHC to output the Core by adding the -ddump-simpl
ﬂag

$ ghc -O --make Sum.hs -ddump-simpl

A ﬁrst taste of Core, part 1

Sum.sum :: [GHC.Types.Int] -> GHC.Types.Int
GblId
[Arity 1
Worker Sum.$wsum
NoCafRefs
Str: DmdType Sm]
Sum.sum =
__inline_me ( (w_sgJ :: [GHC.Types.Int]) ->
case Sum.$wsum w_sgJ of ww_sgM { __DEFAULT ->
GHC.Types.I# ww_sgM
})

A ﬁrst taste of Core, part 2
Rec {
Sum.$wsum :: [GHC.Types.Int] -> GHC.Prim.Int#
GblId
[Arity 1
NoCafRefs
Str: DmdType S]
Sum.$wsum = (w_sgJ :: [GHC.Types.Int]) ->
case w_sgJ of _ {
[] -> 0;
: x_ade xs_adf ->
case x_ade of _ { GHC.Types.I# x1_agv ->
case Sum.$wsum xs_adf of ww_sgM { __DEFAULT ->
GHC.Prim.+# x1_agv ww_sgM
}
}
}
end Rec }

Reading Core: a guide

All names a fully qualiﬁed (e.g. GHC.Types.Int instead of just
Int)
The parts after the function type declaration and before the
function deﬁnition are annotations e.g.
GblId
[ Arity ...]
We’ll ignore those for now
Lots of the names are generated by GHC (e.g. w sgJ).
Note: The Core syntax changes slightly with new compiler releases.

Tips for reading Core

Three tips for reading Core:
Open and edit it in your favorite editor to simplify it (e.g.
rename variables to something sensible).
Use the ghc-core package on Hackage
Use the GHC Core major mode in Emacs (ships with
haskell-mode)

Cleaned up Core for sum

$wsum :: [Int] -> Int#
$wsum = (xs :: [Int]) ->
case xs of _
[] -> 0
: x xs’ -> case x of _
I# x# -> case $wsum xs’ of n#
__DEFAULT -> +# x# n#

sum :: [Int] -> Int
sum = __inline_me ( (xs :: [Int]) ->
case $wsum xs of n#
__DEFAULT -> I# n#)

Core for sum explained

Convention: A variable name that ends with # stands for an
unboxed value.
GHC has split sum into two parts: a wrapper, sum, and a
worker, $wsum.
The worker returns an unboxed integer, which the wrapper
wraps in an I# constructor.
+# is addition for unboxed integers (i.e. a single assembler
instruction).
The worker is not tail recursive, as it performs an addition
after calling itself recursively.
GHC has added a note that sum should be inlined.

Core, case, and evaluation

In core, case always means “evaluate”. Read

case xs of _
[] -> ...
: x xs’ -> ...

as: evaluate xs and then evaluate one of the branches.
Case statements are the only place where evaluation takes
place.
Except when working with unboxed types where e.g.
f :: Int# -> ...
f (x# +# n#)
should be read as: evaluate x# +# n# and then apply the
function f to the result.

Tail recursive sum

module TRSum where

import Prelude h i d i n g (sum)

sum : : [ I n t ] −> I n t
sum = sum ’ 0
where
sum ’ a c c [ ] = acc
sum ’ a c c ( x : x s ) = sum ’ ( a c c + x ) x s
Compiling:

$ ghc -O --make TRSum.hs -ddump-simpl

Core for tail recursive sum
sum1 :: Int; sum1 = I# 0

$wsum’ :: Int# -> [Int] -> Int#
$wsum’ = (acc# :: Int#) (xs :: [Int]) ->
case xs of _
[] -> acc#
: x xs’ -> case x of _
I# x# -> $wsum’ (+# acc# x#) xs’

sum_sum’ :: Int -> [Int] -> Int
sum_sum’ = __inline_me ( (acc :: Int) (xs :: [Int]) ->
case acc of _
I# acc# -> case $wsum’ acc# xs of n#
__DEFAULT -> I# n#)

sum :: [Int] -> Int
sum = __inline_me (sum_sum’ sum1)

Core for tail recursive sum explained

$wsum’s ﬁrst argument is an unboxed integer, which will be
passed in a register if one is available.
The recursive call to $wsum is now tail recursive.

Polymorphic functions

In Core, polymorphic functions get an extra argument for each
type parameter e.g.
i d : : a −> a
id x = x
becomes

id :: forall a. a -> a
id = (@ a) (x :: a) -> x

This parameter is used by the type system and can be ignored.

Exercise

Compare the Core generated by foldl and foldl’ and ﬁnd where they
diﬀer. Start with:
module F o l d C o r e where


foldl = . . .
foldl ’ = . . .
Compile with:

$ ghc -O --make FoldCore.hs -ddump-simpl

Solution

The diﬀerence is in the recursive call:
foldl ’:
: x xs ->
case f z x of z’
__DEFAULT -> foldl’ f z’

foldl :
: x xs -> foldl f (f z x) xs

Profiling

Profiling lets us find performance hotspots.
Typically used when you already have performance problems.

Using proﬁling to ﬁnd a space leak

import System . E n v i r o n m e n t

main = do
[ d ] <− map read ‘ fmap ‘ getArgs
p r i n t ( mean [ 1 . . d ] )

mean x s = sum x s / f r o m I n t e g r a l ( l e n g t h x s )
Compiling:

$ ghc -O --make SpaceLeak.hs -rtsopts

Simple garbage collector statistics

$ ./SpaceLeak 1e7 +RTS -sstderr
./SpaceLeak 1e7 +RTS -sstderr
5000000.5
763,231,552 bytes allocated in the heap
656,658,372 bytes copied during GC
210,697,724 bytes maximum residency (9 sample(s))
3,811,244 bytes maximum slop
412 MB total memory in use (3 MB lost due
to fragmentation)

Generation 0: 1447 collections, 0 parallel,
0.43s, 0.44s elapsed
Generation 1: 9 collections, 0 parallel,
0.77s, 1.12s elapsed

Simple garbage collector statistics cont.

INIT time 0.00s ( 0.00s elapsed)
MUT time 0.54s ( 0.54s elapsed)
GC time 1.20s ( 1.56s elapsed)
EXIT time 0.00s ( 0.00s elapsed)
Total time 1.74s ( 2.10s elapsed)

%GC time 69.0% (74.1% elapsed)

Alloc rate 1,417,605,200 bytes per MUT second

Productivity 30.9% of total user, 25.6% of total
elapsed

The important bits

The program used a maximum of 412 MB of heap
The were 1447 minor collections
The were 9 major collections
69% of the time was spent doing garbage collection!

Time proﬁling

$ ghc -O --make SpaceLeak.hs -prof -auto-all -caf-all
-fforce-recomp
$ ./SpaceLeak 1e7 +RTS -p

Lets increase the stack size to make the program ﬁnish

$ ./SpaceLeak 1e7 +RTS -p -K400M

Time proﬁling results

COST CENTRE MODULE %time %alloc

CAF:main_sum Main 81.0 29.6
mean Main 14.3 0.0
CAF GHC.Float 4.8 70.4

The information isn’t conclusive, but it seems like the runtime is
due to the sum function and the allocation of lots of Doubles.

Space proﬁling

Lets take another perspective by proﬁling the program space usage:

$ ./SpaceLeak 1e7 +RTS -p -K400M -hy

Generate a pretty graph of the result:

$ hp2ps -c SpaceLeak.hp

Space usage graph

SpaceLeak 1e7 +RTS -p -K400M -hy
bytes 8,896,936,338 bytes x seconds Sat Sep 25 14:27 2010

350M

[]

300M

250M
Double

200M

*
150M

100M

BLACKHOLE

50M

0M
0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 seconds

Space leak explained

From the graph we can conﬁrm that a list of Doubles are being
allocated but not freed immediately. The culprit is our formulation
of mean
mean x s = sum x s / f r o m I n t e g r a l ( l e n g t h x s )
sum causes the list xs to be generated, but the list elements
cannot be freed immediately as it’s needed later by length.

Space leak ﬁx

We’ve already seen the solution: compute the sum and the length
at the same time:
where
( s , l ) = f o l d l ’ s t e p (0 , 0) xs
s t e p ( ! s , ! l ) a = ( s+a , l +1)

Writing high-performance code

1. Profile to find performance hotspots
2. Write a benchmark for the offending function
3. Improve the performance, by adjusting evaluation order and
unboxing
4. Look at the generated Core
5. Run benchmarks to confirm speedup
6. Repeat

References / more content

The Haskell Wiki:
http://www.haskell.org/haskellwiki/Performance
Real World Haskell, chapter 25
My blog: http://blog.johantibell.com/

Bonus: Argument unboxing

Strict function arguments are great for performance.
GHC can often represents these as unboxed values, passed
around in registers.
GHC can often infer which arguments are stricts, but can
sometimes need a little help.

Making functions stricter

As we saw earlier, insert is not strict in its ﬁrst argument.
data BST = L e a f | Node I n t BST BST

i n s e r t : : I n t −> BST −> BST
insert x Leaf = Node x L e a f L e a f
The ﬁrst argument is represent as a boxed Int value.

Core for insert

insert :: Int -> BST -> BST
insert = (x :: Int) (t :: BST) ->
case t of _
Leaf -> Node x Leaf Leaf
Node x’ l r ->
case x of _
I# x# -> case x’ of _
I# x’# -> case <# x# x’# of _
False -> case ># x# x’# of _
False -> Node x l r
True -> Node x’ l (insert x r)
True -> Node x’ (insert x l) r

Making functions stricter, part 2

We can make insert strict in the ﬁrsst argument by using a bang
pattern.
{−# LANGUAGE B a n g P a t t e r n s #−}

i n s e r t : : a −> BST a −> BST a
i n s e r t ! x L e a f = Node x L e a f L e a f
Now both equations always cause x to be evaluated.

Core for stricter insert

$winsert :: Int# -> BST -> BST
$winsert = (x# :: Int#) (t :: BST) ->
case t of _
Leaf -> Node (I# x#) Leaf Leaf
Node x’ l r -> case x’ of _
I# x’# -> case <# x# x’# of _
False -> case ># x# x’# of _
False -> Node (I# x#) l r
True -> Node x’ l ($winsert x# r)
True -> Node x’ ($winsert x# l) r

Beneﬁts of strict functions

There are two main beneﬁts of to making functions strict:
No indirections when accessing arguments
Less allocation

High-Performance Haskell

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