# Booleans

Today we’re going to add booleans to our interpreter and our compiler. Specifically, we’re going to add support for these expressions:

`true`

and`false`

, the two boolean values`(not e)`

, a unary operation which evaluates to`true`

on the boolean value`false`

and`false`

otherwise`(num? e)`

, a unary opertion which evaluates to`true`

if`e`

is a number and`false`

otherwise`(zero? e)`

, a unary opertion which evaluates to`true`

if`e`

is the number`0`

and`false`

otherwise

## Types in the interpreter

Up until now, our language has supported only one type: numbers. This means that:

- The values produced by all expressions are numbers
- The values accepted by all operations are numbers

For instance, `add1`

is an operation that takes in a number and produces a
number.

Now we’re going to add booleans to our language. This means adding some expressions that produce booleans, and also some operations that accept booleans.

Here’s our `interp_exp`

function from last time:

let rec interp_exp (exp : s_exp) : int = match exp with | Num n -> n | Lst [Sym "add1"; arg] -> (interp_exp arg) + 1 | Lst [Sym "sub1"; arg] -> (interp_exp arg) - 1 | _ -> raise (BadExpression exp)

Since all of the expressions in our language evaluate to integers, `interp_exp`

can return an integer. How will we modify our interpreter to work with booleans?

One option would be to represent booleans as numbers. For instance, we could
decide that `true`

is 1 and `false`

is 0. Then we could implement our operations
like this:

let rec interp_exp (exp : s_exp) : int = match exp with | Num n -> n | Sym "true" -> 1 | Sym "false" -> 0 | Lst [Sym "add1"; arg] -> (interp_exp arg) + 1 | Lst [Sym "sub1"; arg] -> (interp_exp arg) - 1 | Lst [Sym "not"; arg] -> if (interp_exp arg) = 0 then 1 else 0 | _ -> raise (BadExpression exp)

This is a perfectly valid approach–it’s more or less how C encodes
booleans. It’s not going to work very well, though, once we have more complex
types like strings and lists (though I guess we could use GĂ¶del numbering if we
really had to). We’ll also have a hard time correctly implementing our `num`

operator. What should `(num true)`

return?

Instead of encoding booleans as numbers, we’re going to introduce a new type:
`value`

. (We’ll also take this opportunity to move our interpreter into its own
file, `interp.ml`

).

open S_exp type value = Number of int | Boolean of bool

Our `interp_exp`

function should return this type. First, we’ll just modify it
to support the same operations it did before:

exception BadExpression of s_exp let rec interp_exp (exp : s_exp) : value = match exp with | Num n -> Number n | Lst [Sym "add1"; arg] as e -> ( match interp_exp arg with | Number n -> Number (n + 1) | _ -> raise (BadExpression e) ) | Lst [Sym "sub1"; arg] -> ( match interp_exp arg with | Number n -> Number (n - 1) | _ -> raise (BadExpression e) ) | e -> raise (BadExpression e)

Notice what we’re doing in the `add1`

and `sub1`

cases: if their argument
doesn’t evaluate to a number, it’s not a valid expression. So, for instance,
`(add1 false)`

won’t evaluate to anything.

Now we can add booleans:

let rec interp_exp (exp : s_exp) : value = match exp with | Num n -> Number n | Sym "true" -> Boolean true | Sym "false" -> Boolean false | Lst [Sym "add1"; arg] as e -> ( match interp_exp arg with | Number n -> Number (n + 1) | _ -> raise (BadExpression e) ) | Lst [Sym "sub1"; arg] as e -> ( match interp_exp arg with | Number n -> Number (n - 1) | _ -> raise (BadExpression e) ) | Lst [Sym "not"; arg] -> if interp_exp arg = Boolean false then Boolean true else Boolean false | Lst [Sym "zero?"; arg] -> if interp_exp arg = (Number 0) then Boolean true else Boolean false | Lst [Sym "num?"; arg] -> ( match interp_exp arg with | Number _ -> Boolean true | _ -> Boolean false ) | e -> raise (BadExpression e)

Notice that our new operations can take in arguments of any type. The Lisp-like language we’re implementing, like Python or Racket or Javascript, is dynamically typed.

Finally, we’ll patch up our top-level interpreter function:

let string_of_value (v : value) : string = match v with | Number n -> string_of_int n | Boolean b -> if b then "true" else "false" let interp (program : string) : string = parse program |> interp_exp |> string_of_value

## Types in the compiler

Now that we have an interpreter to test against, we can extend our compiler to support our new operations!

When our interpreter is executing a program, values of expressions are instances
of the `value`

datatype we just defined. We won’t be able to do that in the
compiler–we can’t define new datatypes in x86-64! Remember that when our
program is executing, its values live in registers (actually, just
`rax`

). Registers store 64-bit integers. Right now the values in our program are
all integers, so this works fine. But how will we add booleans? Take a second
and think about how you might implement this.

Well, we know that all of our values need to be represented, at runtime, as 64-bit integers. So instead of representing integers as themselves:

0 -> 0b00 1 -> 0b01 2 -> 0b10 3 -> 0b11 ...

We’re going to represent the integer `x`

as `x << 2`

(shifted left by two bits):

0 -> 0b0000 1 -> 0b0100 2 -> 0b1000 3 -> 0b1100

This is exactly equivalent to representing each integer `x`

as `x * 4`

.

This means our integers have to fit in 62 bits instead of 64. So our minimum
integer is now `-(2**61)`

and our maximum integer is `(2**61) - 1`

.

This also means there are a bunch of 64-bit integers (how many?) that are no
longer being used to represent values! All of our integer values now end with
`00`

. So anything that ends with a different pair of bits won’t be used to
represent a number. This means we can use some of them to represent booleans,
and other types!

First, though, let’s update our compiler to use this new representation for
integers. Integer constants will be easy–we’ll just shift them left. How will
we handle `add1`

and `sub1`

? Well, remember that our runtime representations are
the values multiplied by 4. Since multiplication distributes over addition (and
subtraction), we can just add (or subtract) 4 instead of 1! So:

let num_shift = 2 let num_mask = 0b11 let num_tag = 0b00 let rec compile_exp (exp : s_exp) : directive list = match exp with | Num n -> [Mov (Reg Rax, Imm (n lsl num_shift))] | Lst [Sym "add1"; arg] -> compile_exp arg @ [Add (Reg Rax, Imm (1 lsl num_shift))] | Lst [Sym "sub1"; arg] -> compile_exp arg @ [Sub (Reg Rax, Imm (1 lsl num_shift))] | e -> raise (BadExpression e)

(`lsl`

is “logical shift left.” We could also just multiply by 4, but it’s
clearer this way.)

What happens if we run a program now?

```
>>> compile_and_run "(add1 4)"
20
```

This makes sense–we’re printing out the runtime representation! We’ll need to fix that. We’ll edit our C runtime:

#include <stdio.h> #include <inttypes.h> #define num_shift 2 #define num_mask 0b11 #define num_tag 0b00 extern uint64_t entry(); void print_value(uint64_t value) { if ((value & num_mask) == num_tag) { int64_t ivalue = (int64_t)value; printf("%" PRIi64, ivalue >> num_shift); } else { printf("BAD VALUE %" PRIu64, value); } } int main(int argc, char **argv) { print_value(entry()); return 0; }