Binary operations
So far, the programs in our language have only had to deal with one value at a
time. That’s quite intentional–by restricting our language in this way we’ve
been able to compile everything using only the rax
register! Today, that
changes. Instead of dealing with one value, we’re going to introduce operations
that deal with–get this--two values. As it turns out, this is much more
challenging!
Binary operations in the interpreter
…Or at least, it’s much more challenging in the compiler. Binary operations in our interpreter really won’t be very different from unary operations! First off, here are the operations we will support:
(+ e1 e2)
adds the values of the expressionse1
ande2
.e1
ande2
should evaluate to numbers.(- e1 e2)
subtracts the values of the expressionse1
ande2
.e1
ande2
should evaluate to numbers.(= e1 e2)
evaluates totrue
ife1
ande2
evaluate to the same value andfalse
otherwise(< e1 e2)
evaluates totrue
ife1
evaluates to a smaller number thane2
andfalse
otherwise.e1
ande2
should evaluate to numbers.
Here are the cases we’ll need to add to interp_exp
:
let interp_exp exp = match exp with (* some cases elided... *) | Lst [Sym "+"; e1; e2] -> ( match (interp_exp e1, interp_exp e2) with | Number n1, Number n2 -> Number (n1 + n2) | _ -> raise (BadExpression exp) ) | Lst [Sym "-"; e1; e2] -> ( match (interp_exp e1, interp_exp e2) with | Number n1, Number n2 -> Number (n1 - n2) | _ -> raise (BadExpression exp) ) | Lst [Sym "="; e1; e2] -> Boolean (interp_exp e1 = interp_exp e2) | Lst [Sym "<"; e1; e2] -> ( match (interp_exp e1, interp_exp e2) with | Number n1, Number n2 -> Boolean (n1 < n2) | _ -> raise (BadExpression exp) )
Notice that this code enforces type-correctness: +
and <
will only work on
numbers. Just as we’ve seen with unary operations and conditionals, the
interpreter is just relying on OCaml’s implementations of these features.
Binary operations in the compiler
Here’s where it gets tricky.
Let’s try to sort of “naively” translate the interpreter version of +
(reminder: right now the compiler, unlike the interpreter, does not enforce
type-correctness):
let compile_exp exp = match exp with (* some cases elided... *) | Lst [Sym "+"; e1; e2] -> compile_exp e1 @ compile_exp e2 @ ...
Remember: compile_exp
emits instructions to compute the value of exp
and to
store that value in rax
. So by the time we want to add the two values, e2
is
going to be in rax
and e1
is going to be lost! So, we’ll somehow need to
“save” the value of e1
. Here’s an idea: we could save it to a different
register! x86-64 has 16 general-purpose registers; let’s use r8
, written Reg
R8
in our OCaml assembly library:
let compile_exp exp = match exp with (* some cases elided... *) | Lst [Sym "+"; e1; e2] -> compile_exp e1 @ [Mov (Reg R8, Reg Rax)] @ compile_exp e2 @ [Add (Reg Rax, Reg R8)]
Here we’re saving compiling e1
into rax
, saving rax
into r8
, compiling
e2
into rax
, then adding the results of the two expressions together. This
seems to work great!
$ compile_and_run "(+ 1 2)";; 3 $ compile_and_run "(+ (+ 1 2) 3)";; 6 $ compile_and_run "(+ 1 (+ 2 3))";; 7
Wait, what was that last result? Something’s not right here. Let’s look at the assembly we’re producing:
entry: mov rax, 4 mov r8, rax mov rax, 8 mov r8, rax mov rax, 12 add rax, r8 add rax, r8 ret
We’re compiling (+ 1 (+ 2 3))
by first storing the runtime representation of
1
in r8
, then compiling the second argument to +
. But since the second
argument is also a call to +
, the first thing it’s going to do is do overwrite
the value in r8
(in this case, with the runtime representation of 2
).
We could try to get around this by using more registers. We could imagine having
our compiler take a list of registers it’s not allowed to use when compiling an
expression–here, since r8
is being used to store 1
, we couldn’t use r8
when compiling (+ 2 3)
. If we had an infinite number of registers, a scheme
like this could work. But since we only have 16, there are going to be
expressions that we won’t be able to compile with that kind of scheme.
So we need someplace to store intermediate values during computation, where we won’t run out of room. How about memory?
The stack
The region of memory that our program has available for temporary use during computations is called the stack. (Longer-lived values live in the heap, which we’ll talk about in a few lectures.) We’ll start with a simple model of this region of memory; we’ll make this model more complex, and somewhat more accurate, when we talk about functions.
Imagine the stack as an array of cells, each of which has an address. The bottom
of our stack is at the highest address. When our program starts executing, the
register rsp
contains this address. The memory cell at this address contains
the function’s return address. We’ll learn more about what that means later; for
now, just know that we shouldn’t overwrite the data at that address.
The “next” memory cell in the stack–that is, the first cell that we can write
data into–is at (rsp - 8)
. Why -
? Because the stack grows “up”, from higher
addresses to lower addresses. rsp + 8
probably contains data used by the
calling function. Why 8
? Because the word size on x86-64 is 8 bytes (64
bits). x86-64 memory addresses are 8 bytes; x86-64 registers are 8 bytes; all of
our program values are 8 bytes. So the stack looks like this:
address | data |
---|---|
… | … |
rsp - 16 |
unused |
rsp - 8 |
unused |
rsp |
address of caller stack frame |
Accessing the stack from assembly
We’ve seen the mov
instruction before–it lets us move immediate data into
registers, or move data between registers. It also lets us move data between
registers and memory. So, let’s modify our compiler to save the value of the
first argument to +
to memory instead of r8
.
let compile_exp exp = match exp with (* some cases elided... *) | Lst [Sym "+"; e1; e2] -> compile_exp e1 @ [Mov (MemOffset (Reg Rsp, Imm (-8)), Reg Rax)] @ compile_exp e2 @ [Mov (Reg R8, MemOffset (Reg Rsp, Imm (-8)))] @ [Add (Reg Rax, Reg R8)]
If we compile (+ 1 2)
now, we get this:
entry: mov rax, 4 mov [rsp + -8], rax mov rax, 8 mov r8, [rsp + -8] add rax, r8 ret
Those square-bracketed expressions are how our assembly language represents
memory accesses. As we see, offsets into memory (of the form <operand> +
<operand>
) can be used as operands to instructions like mov
and add
.
What happens if we compile (+ 1 (+ 2 3))
now? We still have the same
problem we did before--2
is overwriting 1
, this time at [rsp - 8]
instead
of in r8
:
entry: mov rax, 4 mov [rsp + -8], rax mov rax, 8 mov [rsp + -8], rax mov rax, 12 mov r8, [rsp + -8] add rax, r8 mov r8, [rsp + -8] add rax, r8 ret
Now, though, we’ll be able to fix this issue.
Tracking the stack index
Instead of storing the intermediate value 2
at [rsp - 8]
, the compiler
should store it at the next available stack address: [rsp - 16]
. So when we
call compile_exp e2
, we will need to let it know that [rsp - 16]
is the new
first stack address.
We can implement this by adding an argument to compile_exp
representing the
next available stack index:
let compile_exp (stack_index : int) exp = ...
Most of the time, this stack_index
argument will remain unchanged through
recursive calls. But if we store something on the stack, we’ll need to update
it. Right now, we need to do that in exactly one place: that compile_exp e2
call. We’ll modify our code to store the intermediate value at [rsp +
stack_index]
, and to subtract 8 from the stack index for that recursive call:
let compile_exp exp = match exp with (* some cases elided... *) | Lst [Sym "+"; e1; e2] -> compile_exp stack_index e1 @ [Mov (MemOffset (Reg Rsp, Imm stack_index), Reg Rax)] @ compile_exp (stack_index - 8) e2 @ [Mov (Reg R8, MemOffset (Reg Rsp, Imm stack_index))] @ [Add (Reg Rax, Reg R8)]
We now get the following code for (+ 1 (+ 2 3))
:
entry: mov rax, 4 mov [rsp + -8], rax mov rax, 8 mov [rsp + -16], rax mov rax, 12 mov r8, [rsp + -16] add rax, r8 mov r8, [rsp + -8] add rax, r8 ret
This now works great! We’ve successfully implemented addition.
Other binary operations
Our code for the other binary operations we support looks similar:
let compile_exp exp = match exp with (* some cases elided... *) | Lst [Sym "-"; e1; e2] -> compile_exp stack_index e1 @ [Mov (MemOffset (Reg Rsp, Imm stack_index), Reg Rax)] @ compile_exp (stack_index - 8) e2 @ [Mov (Reg R8, Reg Rax)] @ [Mov (Reg Rax, MemOffset (Reg Rsp, Imm stack_index))] @ [Sub (Reg Rax, Reg R8)] | Lst [Sym "="; e1; e2] -> compile_exp stack_index e1 @ [Mov (MemOffset (Reg Rsp, Imm stack_index), Reg Rax)] @ compile_exp (stack_index - 8) e2 @ [Mov (Reg R8, MemOffset (Reg Rsp, Imm stack_index))] @ [Cmp (Reg Rax, Reg R8)] @ zf_to_bool | Lst [Sym "<"; e1; e2] -> compile_exp stack_index e1 @ [Mov (MemOffset (Reg Rsp, Imm stack_index), Reg Rax)] @ compile_exp (stack_index - 8) e2 @ [Mov (Reg R8, MemOffset (Reg Rsp, Imm stack_index))] @ [Cmp (Reg R8, Reg Rax)] @ lf_to_bool
<
uses lf_to_bool
, which calls setl
instead of setz
. setl
reads the
SF
and OF
flags; after a comparison operation, it will set its operand to
1
if the first comparison argument was strictly less than the second.
A note about undefined behavior
What should this expression evaluate to?
(+ 1 false)
Our interpreter gives us the answer: just like a nonsense expression like
(hello csci1260)
, (+ 1 false)
isn’t part of our language, so the interpreter
raises an exception. What will our compiler do on this program?
$ compile_and_run "(+ 1 false)";; "BAD 35"
Our runtime indicates that we’ve produced a bad value (specifically, 35)–it doesn’t correspond to anything in our tagging scheme. So, OK–the compiler and the interpreter both end up producing errors on this program.
How about this program?
(+ 32 false)
Our interpreter, of course, still throws an exception. But our compiler does something pretty weird:
$ compile_and_run "(+ 32 false)";; "true"
Weird, right? It makes sense, though: since false
is represented as
0b00011111
and 32
is represented as 0b10000000
, false + 32
is
0b10011111
, the runtime representation of true
.
So: is this a bug in our compiler? Maybe not! (+ 32 false)
isn’t a valid
program in the language supported by our compiler. There are lots of these
invalid programs! Different ones result in different things:
- Programs like
(hello csci1260)
result in an exception at compile time - Programs like
(+ 1 false)
result in a runtime error - Programs like
(+ 32 false)
result in a strange value
The behavior of our compiler on these programs is undefined. We can error our at compile-time, error out at runtime, produce a reasonable-looking value, or anything else. Some real-world programming languages include undefined behavior as part of the language standard; for instance, dereferencing a null pointer in C is undefined.
Many modern languages, however, eschew undefined behavior–as we have just seen,
it’s quite confusing for programmers! A reasonable specification for programs
like (+ 1 true)
is that they should result in errors. In a couple of weeks, we
will see how to add error-handling to our compiler and get rid of this strange
behavior. For now, we’ll leave our compiler’s behavior specified only on valid
expressions.