# Implementing Laziness in C

The aim of this blog post is to explain haskell's (specifically, GHC) evaluation model without having to jump through too many hoops. I'll explain how pervasive non-strictness is when it comes to Haskell, and why compiling non-strictness is an interesting problem.

# Showing off non-strictness:

We first need a toy example to work with to explain the fundamentals of non-strict evaluation, so let's consider the example below. I'll explain the `case`

construct which is explained below. Don't worry if ato lazy evaluation "looks weird", it feels that way to everyone till one plays around with it for a while!

We will interpret this example as both a strict program and a non-strict program. This will show us that we obtain different outputs on applying different interpretations.

We distinguish between primitive values (integers such as `1`

, `2`

, `3`

) and boxed values (functions, data structures). Boxed values can be evaluated non-stricly. Primitive values do not need evaluation: they are primitive.

##### Code

```
-- Lines starting with a `--` are comments.
-- K is a function that takes two arguments, `x` and `y`, that are both boxed values.
-- K returns the first argument, `x`, ignoring the second argument, `y`.
-- Fun fact: K comes for the K combinator in lambda calculus.
kCombinator :: Boxed -> Boxed -> Boxed
kCombinator x y = x
-- one is a function that returns the value 1# (primitive 1)
one :: PrimInt
one = 1
-- Loopy is a function that takes zero arguments, and tries to return a boxed value.
-- Loopy invokes itself, resulting in an infinite loop, so it does not actually return.
loopy :: Boxed
loopy = loopy
-- main is our program entry point.
-- main takes no arguments, and returns nothing
main :: Void
main = case kCombinator one loopy of -- force K to be evaluated with a `case`
kret -> case kret of -- Force evaluation
i -> printPrimInt i -- Call the forced value of `kret` as `i` and then print it.
```

##### A quick explanation about `case`

:

`case`

is a language construct that *forces* evaluation. In general, no value is evaluated unless it is *forced* by a case.

# Analysing the example:

###### The strict interpretation:

If we were coming from a strict world, we would have assumed that the expression `K one loopy`

would first try to evaluate the arguments, `one`

and `loopy`

. Evaluating `one`

would return the primitive value `1`

, so this has no problem.

On trying to evaluate `loopy`

, we would need to re-evaluate `loopy`

, and so on ad infitum, which would cause this program to never halt.

This is because, as programmers coming from a strict world, we assume that *values are evaluated as soon as possible*.

So, the output of this program is to have the program infinite-loop for ever, under the strict interpretation.

###### The non-strict interpretation:

In the non-strict world, we try to evaluate `K(1, loopy)`

since we are asked the result of it by the `case`

expression. However, we do not try to evaluate `loopy`

, since no one has asked what it's value is!

Now, we know that

```
kCombinator x y = x
```

Therefore,

```
kCombinator one loopy = one
```

regardless of what value `loopy`

held.

So, at the case expression:

```
main = case K(one, loopy) of -- force K to be evaluated with a `case`
>>> kret -> ...
```

`kret = one`

, we can continue with the computation.

```
main :: () -> Void
main = case kCombinator one loopy of -- force K to be evaluated with a `case`
kret -> case kret of -- Call the return value of K as `x`, and force evaluation.
>>> i -> printPrimInt i -- Call the vreturn value of `x` as `primx` and then print it.
```

Here, we force `kret`

(which has value `one`

) to be evaluated with `case kret of...`

. since `one = 1`

, `i`

is bound to the value `1`

. Once `i`

is returned, we print it out with `printPrimInt(primx)`

.

The output of the program under non-strict interpretation is for it to print out `1`

.

# Where does the difference come from?

Clearly, there is a divide: strict evaluation tells us that this program should never halt. Non-strict evaluation tells us that this program will print an output!

To formalize a notion of strictness, we need a notion of `bottom`

(`_|_`

).

A value is said to be `bottom`

if in trying to evaluate it, we reach an undefined state. (TODO: refine this, ask ben).

Now, if a function is *strict*, it would first evaluate its arguments and then compute the result. So, if a strict function is given a value that is `bottom`

, the function will try to evaluate the argument, resulting in the computation screwing up, causing the output of the whole function to be `bottom`

.

Formally, a function `f`

is strict iff (if and only if) `f(bottom) = bottom`

.

Conversely, a *non-strict* function does not need to evaluate its arguments if it does not use them, as in the case of `K 1 loopy`

. In this case, `f(bottom)`

need not be equal to bottom.

Formally, a function `f`

is non-strict iff (if and only if) `f(bottom) /= bottom`

.

As Paul Halmos says, " A good stack of examples, as large as possible, is indispensable for a thorough understanding of any concept, and when I want to learn something new, I make it my first job to build one.". Let us consider some examples.

`id`

```
id x = x
id (3) = 1
id (bottom) = bottom
```

`id`

is strict, since `id(bottom) = bottom`

.

`const`

```
const_one x = 1
const_one(bottom) = 1
const_one(3) = 1
```

`const_one`

is not strict, as `const_one(bottom) /= bottom`

.

`K`

```
K x y = x
K 1 2 = 1
K 1 bottom = 1
K bottom 2 = bottom
```

Note that `K(bottom, y) = bottom`

, so K is *strict in its first argument*, and `K(x, bottom) /= bottom`

, so K is *non-strict in its second argument*.

This is a neat example showing how a function can be strict and lazy in different arguments of the function.

# Compiling non-strictness, v1:

## How does GHC compile non-strictness?

`GHC`

(the Glasgow haskell compiler) internally uses multiple intermediate representations, in order of original, to what is finally produced:

- Haskell (the source language)
- Core (a minimal set of constructs to represent the source language)
- STG (Spineless tagless G-machine, a low-level intermediate representation that accurately captures non-strict evaluation)
- Câ€“ (A C-like language with GHC-specific customization to support platform independentcode generation).
- Assembly

Here, I will show how to lower simple non-strict programs from a **fictitous** `Core`

like language down to `C`

, while skipping `STG`

, since it doesn't really add anything to the high-level discussion at this point.

## Our example of compiling non-strictness

Now, we need a *strategy* to compile the non-strict version of our program. Clearly, `C`

cannot express laziness directly, so we need some other mechanism to implement this. I will first code-dump, and then explain as we go along.

###### Executable `repl.it`

:

###### Source code

```
#include <assert.h>
#include <stdio.h>
/* a boxed value is a function that can be executed to compute something.
* We make the return value `void` on purpose. This needs to be typecast to a concrete
* Boxed type to get a value out of it: eg, typecast to BoxedInt.
*/
typedef void (*Boxed)();
/* A boxed int, that on evaluation yields an int*/
typedef int (*BoxedInt)();
/* one = 1# */
int one() {
return 1;
}
/* bottom = bottom */
void bottom() {
printf("in function: %s\n", __FUNCTION__);
bottom();
}
/* K x y = x */
Boxed K(Boxed x, Boxed y) {
return x;
}
/*
main :: () -> Void
main = case K(one, loopy) of -- force K to be evaluated with a `case`
kret -> case kret of -- Call the return value of K as `x`, and force evaluation.
i -> printPrimInt(i) -- Call the vreturn value of `x` as `primx` and then print it.
*/
int main() {
Boxed kret = K((Boxed)one, (Boxed)bottom);
int i = (*(BoxedInt)kret)();
printf("%d", i);
return 1;
}
```

We convert every possibly lazy value into a `Boxed`

value, which is a function pointer that knows how to compute the underlying value. When the lazy value is forced by a `case`

, we call the `Boxed`

function to compute the output.

This is a straightforward way to encode non-strictness in C. However, do note that this is not *lazy*, because a value could get recomputed many times. *Laziness* guarantees that a value is only computed once and is later memoized.

# Compiling with a custom call stack / continuations

As one may notice, we currenly use the native call stack every time we *force* a lazy value. However, in doing so, we might actually run out of stack space, which is undesirable. Haskell programs like to have "deep" chains of values being forced, so we would quite likely run out of stack space.

Therefore, GHC opts to manage its own call stack on the heap. The generated code looks as you would imagine: we maintain a stack of function pointers + auxiliary data ( stack saved values), and we push and pop over this "stack". When we run out of space, we `<find correct way to use mmap>`

to increase our "stack" size.

I've played around with this value a little bit, and have found that the modern stack size is quite large: IIRC, It allowed me to allocate ~`26 GB`

. I believe that the amount it lets you allocate is tied directly to the amount of physical memory + swap you have. I'm not too sure, however. So, for my haskell compiler, `sxhc`

, I am considering cheating and just using the stack directly.

Code for the same example (with the K combinator) is provided here.

###### Executable `repl.it`

:

###### Source code

```
#include <assert.h>
#include <stdio.h>
#define STACK_SIZE 50000
/* a boxed value is a function that can be executed to compute something. */
typedef void (*Boxed)();
/* a return continuation that receives a boxed value. */
typedef void (*BoxedContinuation)(Boxed);
/* A return continuation that receives an int value. */
typedef void (*IntContinuation)(int);
/* Custom stack allocated on the .data section*/
void *stack[STACK_SIZE];
/* Stack pointer */
int sp = 0;
/* Push a continuation `cont` */
void pushContinuation(void *cont) {
assert(sp >= 0);
assert(sp < STACK_SIZE);
stack[sp] = cont;
sp++;
}
/* Pop a continuation frame. */
void *popContinuation() {
assert(sp < STACK_SIZE);
assert(sp >= 0);
sp--;
void *cont = stack[sp];
return cont;
}
/* one = 1# */
void one() {
printf("in function: %s\n", __FUNCTION__);
void *f = popContinuation();
(*(IntContinuation)(f))(1);
}
/* bottom = bottom */
void bottom() {
printf("in function: %s\n", __FUNCTION__);
bottom();
}
/* K x y = x */
void K(Boxed x, Boxed y) {
printf("in function: %s\n", __FUNCTION__);
void *f = popContinuation();
(*(BoxedContinuation)(f))(x);
}
void XForceContinuation(int i) {
printf("in function: %s\n", __FUNCTION__);
printf("%d", i);
}
void KContinuation(Boxed x) {
printf("in function: %s\n", __FUNCTION__);
pushContinuation((void *)XForceContinuation);
x();
}
int main() {
printf("in function: %s\n", __FUNCTION__);
pushContinuation((void *)KContinuation);
K(one, bottom);
return 0;
}
```

we maintain our own "call stack" of continuations. These continuations are precisely the parts of the code that deal with the return value of a case. ever

```
case x of
xeval -> expr
```

compiles to:

```
pushContinuation(XEvalContinuation);
x()
```

That is, push a continuation, and then "enter" into `x`

.

One might have a question: does this still not use the call stack? There is a function call at the end of most functions in the source code, so in theory, we *are* using the call stack, right? The answer is no. It's thanks to a neat optimisation technique called *tail call elimination*. The observation is that *after the call*, there is no more code to execute in the caller. So, by playing some stack tricks, one can convert a *call* to a *jump*.

Remember, a * call* instruction uses the stack to setup a stack frame, under the assumption that we will

*at some point. But, clearly, under our compilation model, we will never*

`ret`

`ret`

, simply call more functions. So, we don't *need*the state maintained by a

`call`

. We can simply `jmp`

.# Wrapping up

I hope I've managed to convey the *essence* of how to compile Haskell. I skipped a couple of things:

- haskell data types: sum and product types. These are straightforward, they just compiled to tagged structs.
`let`

bindings: These too are straightforward, but come with certain retrictions in STG. It's nothing groundbreaking,andis well written in the paper.- Black holes: Currently, we are not truly
*lazy*, in that we do not update values once they are computed. - GC: how to weave the GC through the computation is somewhat non-trivial.

All of this is documented in the excellent paper: Implementing lazy languages on stock hardware.

# Appendix: Non-strict versus Lazy

I have use the word `non-strict`

throughout, and not `lazy`

. There is a technical difference between the two, which is that:

`non-strict`

is a evaluation order detail that guarantees that values are`lazy`

is one way to*implement*`non-strict`

, that provides guarantees that a value will not be evaluated twice.

For example, consider the small haskell snippet:

```
f :: Int -> Int -> Int
f x y = x * 2
loop :: a
loop = loop
main :: IO ()
main = let a = 10 + 10 print (f a loop)
```

We expect that the program will *never* evaluate `y=loop`

because `f`

does not use `y`

anywhere. We want the program to return the answer:

```
f 10 loop
= (10 + 10) * (10 + 10)
= 20 * 20
= 400
```

However, we don't really care if the program is evaluated as shown here, using *lazy evaluation*, (where the blue color indicates what is being evaluated):

or it's being evaluated as shown here, where the expression `a`

is evaluated twice:

See that both these programs compute the same answer, but do so differently. In the former case of lazy evaluation the variable `a`

is *shared* across both occurences: it is computed once, and the result is reused. In the latter case of non-strict evaluation, we see that the variable `a`

is *evaluated twice independently*, once for each occurrence of `a`

.

There are trade-offs with both approaches. In the lazy evaluation case, we are guaranteed that a large expression will not be evaluated twice. On the flip side, in the case of non-strict evaluation, since we can evaluate the same expression multiple times, we can *exploit parallelism*: Two cores can conceivably compute `10 + 10`

in parallel to arrive at `20`

. This is in contrast to the lazy evaluation case where this is not allowed, since the compiler must *guarantee* that an expression is evaluated at most once. If the same expression `10 + 10`

is evaluated on two cores (in parallel), it is still evaluated twice! So there is a trade-off between guarantees to the end user (that an expression will not be evaluated twice), and freedom for the compiler (to evaluate in parallel). This is a common theme in language and compiler design, and one I find very interesting.

## Related Jobs

## Related Articles

## Related Issues

- Started
- 0
- 1
- Intermediate

- Started
- 0
- 2
- Intermediate

- Open
- 0
- 0
- Intermediate

- Submitted
- 1
- 1
- Intermediate

### Get hired!

#### Sign up now and apply for roles at companies that interest you.

Engineers who find a new job through WorksHub average a *15%* increase in salary.