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Haskell programs: how do they run?

If this article were entitled “Haskell programs: how do they look?” it would be very short. The answer would be “beautiful”. If it were entitled “Haskell programs: what do they calculate?” I could write about Haskell’s lovely mathematical compositional behaviour.

But it’s not; it’s entitled “Haskell programs: how do they run?” and explains how Haskell programs are executed step by step (specifically, the programs produced by GHC). Haskell is what is called a “lazy language” and because lazy languages are rare the execution of Haskell programs often remains rather mysterious even to its most frequent users. The consequence of this lack of understanding is programs that have strange performance characteristics, often using vastly larger amounts of memory than anticipated, and sometimes running more slowly too.

This article will present a model that you can use to understand the mechanism by which Haskell programs are executed. It’s not complicated, but useful information on this topic is very hard to find and it took me several years to put together all the key pieces of information needed for a complete picture. I’ll reassemble the complete picture for you much more quickly in the space of merely several paragraphs below!

To fit in with the nomenclature in the Haskell literature I’m going to use the word “evaluation” rather than “execution”. The latter seems to better convey the notion of following a sequence of instructions encoded by a program, but the former seems to be in wider use by those who write compilers.

Firstly, let me mention some terms that are often used when talking about how Haskell programs are evaluated: “redex”, “weak head normal form” (WHNF), “constant applicative form” (CAF), “call by need”. As far as I can tell these are all terms from the mathematical theory of lambda calculus and although they describe aspects of the process of evaluation of Haskell programs they are not useful for understanding the process of evaluation directly. Two other phrases that are, in my opinion, unhelpful and imprecise respectively, are “graph reduction” and “delays the evaluation of an expression until its value is needed”.

Normal form

By contrast I’m going to offer a very direct and concrete approach to understanding how Haskell programs are evaluated. The evaluation will be done in terms of something I’m going to call the “normal form”. The normal form is a simplified version of Haskell with fewer language constructs. It has only

Additionally, in a function call the function itself and arguments must be variables or literals; and constructor applications must be “saturated”, i.e. there must be no missing arguments.

All Haskell programs can be translated into this normal form. I’m not going to expend space describing how because there’s lots and lots of Haskell syntax. Suffice it to say that these can all be translated straightforwardly to the normal form: where clauses, function definition syntax, guards, nested patterns, nested expressions.

This normal form is a language with a direct operational reading. You could even call in an imperative language. I’m not going to give a special name to the normal form. It’s similar to “administrative normal form” which is a concept from compiler development technology, and it’s also similar to STG, one of the GHC intermediate languages.

The evaluation of this normal form is sufficient for 99% of your Haskell performance needs. It will be enough to understand any performance issues except the lowest level performance hackery.

Here’s an example of a program in Haskell and its equivalent in the normal form:

-- Haskell
map f []     = []
map f (x:xs) = f x : map f xs

-- Normal form
map = \f xs -> case xs of
  []    -> []
  x:xs' -> let first = f x
               rest  = map f xs'
           in first : rest

Notice that we can’t use Haskell’s multi-clause function definition syntax, the arguments must specifically be in lambdas and we can’t have nested subexpressions; instead we have to bind subexpressions to variables.

What do we evaluate to?

Since we are considering a process for evaluating Haskell programs our first thought ought to be about what we expect from the process: what do Haskell programs evaluate to?

The result of an evaluation is called a “value”, and a value is either

(This is the same concept that is captured by the notion of WHNF.)

How do we evaluate

How exactly does evaluation proceed? It’s actually rather imperative.

Those are the complete rules for evaluating Haskell programs. Some examples will follow shortly.

Resource usage

There are just two simple rules which fully explain how these evaluation rules cause usage of memory resources.

Firstly, the only thing that allocates on the heap is let.

Secondly, the only thing that consumes stack (in a way that we care about) is case, whilst it is evaluating its scrutinee. (The “scrutinee” is s in case s of ... -> ....) Haskell consumes stack whilst evaluating s so it can know where it should continue from once evaluation of s has finished.

Nothing else consumes any memory resources.

Evaluation example

We’ll trace an example implemented in terms of map above plus the following two functions

-- Haskell
repeat x = xs where xs = x : xs

-- Normal form
repeat = \x -> let xs = x : xs
               in xs

-- Haskell
head (x:xs) = x

-- Normal form
head = \xs -> case xs of x:xs' -> x

The expression to evaluate will be

-- Haskell
head (map (\x -> x + x) (repeat (10 + 1)))

-- Normal form
let f = \x -> x + x
    t = 10 + 1
    r = repeat t
    m = map f r
in head m

To save space in this article you can watch my talk at Haskell eXchange 2016 where I trace the evaluation of this program.

Primitive operations

Something to bear in mind is that primitive operations must run on values, i.e. fully evaluated things. This means that an operation such as (+) must be implemented as something like

(+) = \x y -> case x of
     x' -> case y of
         y' -> primitive_plus x' y'

where primitive_plus is perhaps implemented by some underlying C library. The consequence is that (+), or any other primitive operation evaluates its arguments.


The rules for evaluation tell us exactly what we need to know about how sharing happens. Consider these two versions of an enumeration function:

enum1 = zip ns
    where ns = [1..]

enum2 xs = zip ns xs
    where ns = [1..]

If we’re familiar with the concept of “eta equivalence” we might think that these two definitions will be evaluated in the same way. We might think of an analogous definition where two definitions are equivalent and evaluated in the same way:

inc1 x = (+) 1 x

inc2   = (+) 1

But the enumeration examples do not evaluate the same way. Why not? Let’s look at what happens when we convert them into the normal form

let enum1 = let ns = [1..]
            in zip ns

let enum2 = \xs -> let ns = [1..]
                   in zip ns xs

Considering how these evaluate we see that in enum1, ns is shared by all invocations of the function, whereas in enum2, ns is created afresh by each invocation of the function. Sharing ns can lead to large and surprising space leaks.

So we see that careful consideration of how Haskell programs evaluate can shed light on unexpected or surprising behaviour.


foldl is expressed in the normal form as

foldl = \f z xs -> case xs of
  []    -> z
  x:xs' -> let z' = f z x
           in foldl f z' xs'

See my talk at Haskell eXchange 2016 where I trace the evaluation of foldl (+) 0 [1..100]. In short, it builds up the proverbial “long chain of thunks” on the heap before consuming O(n) stack space to evaluate them.


foldl' is the “strict version of foldl” and is expressed in the normal form as

foldl' = \f z xs -> case xs of
  []    -> z
  x:xs' -> case f z x of
      z' -> foldl' f z' xs'

(As a short aside, note that in Haskell we would use seq instead of case because a Haskell case with no patterns doesn’t evaluate its scrutinee.)

See my talk at Haskell eXchange 2016 where I trace the evaluation of foldl' (+) 0 [1..100]. Unlike foldl, it consumes neither (much) heap nor stack.


foldr is expressed in the normal form as

foldr = \f z xs -> case xs of
  []    -> z
  x:xs' -> let rest = foldr (+) 0 xs'
           in f x rest

See my talk at Haskell eXchange 2016 where I trace the evaluation of foldr (+) 0 [1..100]. It consumes O(n) heap and stack.


All Haskell programs can be translated straightforwardly to a simple normal form which has a simple imperative-style interpretation.

By following through the execution of the program we can understand how it uses memory resources.