Next, lets add support for
- Multiple datatypes (
number
andboolean
) - Calling external functions
In the process of doing so, we will learn about
- Tagged Representations
- Calling Conventions
Plan
Our plan will be to (start with boa
) and add the following features:
Representing boolean values (and numbers)
Arithmetic Operations
Arithmetic Comparisons
Dynamic Checking (to ensure operators are well behaved)
1. Representation
Motivation: Why booleans?
In the 21st century, its a bit silly to use
0
forfalse
and- non-zero for
true
.
But really, boolean
is a stepping stone to other data
- Pointers
- Tuples
- Structures
- Closures
The Key Issue
How to distinguish numbers from booleans?
- Need extra information to mark values as
number
orbool
.
A Word
(A reminder for me, since I always get mixed up)
- A Bit is 1-bit
- A Byte is 8-bits
- A Word is 2-bytes = 16 bits
- A Double Word is 2-words = 4-bytes = 32 bits
- A Quad Word is 4-words = 8-bytes = 64 bits
We are working in x86_64
where the default size is a qword
- Registers are 64-bits
- Arithmetic is 64-bits
- Stack slots should be 64-bits
- etc.
Option 1: Use Two (Quad-)Words
How to distinguish numbers from booleans?
Need extra information to mark values as
number
orbool
.
First word is 0
means bool
, is 1
means number
, 2
means pointer etc.
Value | Representation (HEX) |
---|---|
3 |
[0x000000000][0x00000003] |
5 |
[0x000000000][0x00000005] |
12 |
[0x000000000][0x0000000c] |
42 |
[0x000000000][0x0000002a] |
false |
[0x000000001][0x00000000] |
true |
[0x000000001][0x00000001] |
Pros
- Can have lots of different types, but
Cons
Takes up double memory,
Operators
+
,-
require two memory reads.
In short, rather wasteful! We don’t need so many types.
Option 2: Use a Tag Bit
Can distinguish two types with a single bit.
Least Significant Bit (LSB) is
0
fornumber
1
forboolean
Question: why not 0
for boolean
and 1
for number
?
Tag Bit: Numbers
So number
is the binary representation shifted left by 1 bit
- Lowest bit is always
0
- Remaining bits are number’s binary representation
For example, in binary:
Value | Representation (Binary) |
---|---|
3 |
[0b00000110] |
5 |
[0b00001010] |
12 |
[0b00011000] |
42 |
[0b01010100] |
Or in hexadecimal:
Value | Representation (HEX) |
---|---|
3 |
[0x06] |
5 |
[0x0a] |
12 |
[0x18] |
42 |
[0x54] |
Tag Bit: Booleans
Most Significant Bit (MSB) is
1
fortrue
0
forfalse
For example
Value | Representation (Binary) |
---|---|
true |
[0b10000000000000000000000000000001] |
false |
[0b00000000000000000000000000000001] |
Or, in HEX
Value | Representation (HEX) |
---|---|
true |
[0x80000001] |
false |
[0x00000001] |
(eliding the 32/8 zeros in the “most-significant” DWORD
)
Types
Lets extend our source types with boolean
constants
data Expr a
= ...
| Boolean Bool a
Correspondingly, we extend our assembly Arg
(values) with
data Arg
= ...
| HexConst Int
So, our examples become:
Value | Representation (HEX) |
---|---|
Boolean False |
HexConst 0x00000001 |
Boolean True |
HexConst 0x80000001 |
Number 3 |
HexConst 0x00000006 |
Number 5 |
HexConst 0x0000000a |
Number 12 |
HexConst 0x0000000c |
Number 42 |
HexConst 0x0000002a |
Transforms
Next, lets update our implementation
The parse
, anf
and tag
stages are straightforward.
Lets focus on the compile
function.
A TypeClass for Representing Constants
Its convenient to introduce a type class describing Haskell types that can be represented as x86 arguments:
class Repr a where
repr :: a -> Arg
We can now define instances for Int
and Bool
as:
instance Repr Int where
= Const (Data.Bits.shift n 1) -- left-shift `n` by 1
repr n
instance Repr Bool where
False = HexConst 0x00000001
repr True = HexConst 0x80000001 repr
Immediate Values to Arguments
Boolean b
is an immediate value (like Number n
).
Lets extend immArg
that transforms an immediate expression to an x86 argument.
immArg :: Env -> ImmTag -> Arg
Var x _) = ...
immArg (Number n _) = repr n
immArg (Boolean b _) = repr b immArg (
Compiling Constants
Finally, we can easily update the compile
function as:
compileEnv :: Env -> AnfTagE -> Asm
@(Number _ _) = [IMov (Reg RAX) (immArg env e)]
compileEnv _ e@(Boolean _ _) = [IMov (Reg RAX) (immArg env e)] compileEnv _ e
(The other cases remain unchanged.)
Lets run some tests to double check.
QUIZ
What is the result of:
> exec "15" ghci
A. Error
B. 0
C. 15
D. 30
Output Representation
Say what?! Need to update our run-time printer in main.c
void print(int val){
if (val == CONST_TRUE)
"true");
printf(else if (val == CONST_FALSE)
"false");
printf(else // should be a number!
"%d", d >> 1); // shift right to remove tag bit.
printf( }
and now we get:
> exec "15"
ghci15
Can you think of some other tests we should write?
QUIZ
What is the result of
> exec "let x = 15 in x" ghci
A. Error
B. 0
C. 15
D. 30
QUIZ
What is the result of
>>> exec "if 3: 12 else: 49"
>>> exec "if 0: 12 else: 49"
>>> exec "if true: 12 else: 49"
>>> exec "if false: 12 else: 49"
A. Error
B. 0
C. 12
D. 49
Lets go and fix the code so the above do the right thing!
2. Arithmetic Operations
Constants like 2
, 29
, false
are only useful if we can perform computations with them.
First lets see what happens with our arithmetic operators.
QUIZ: Addition
What will be the result of:
> exec "12 + 4" ghci
A. Does not compile
B. Run-time error (e.g. segmentation fault)
C. 16
D. 32
E. 0
Shifted Representation and Addition
We are representing a number n
by shifting it left by 1
n
has the machine representation2*n
Thus, our source values have the following _representations:
Source Value | Representation (DEC) |
---|---|
3 |
6 |
5 |
10 |
3 + 5 = 8 |
6 + 10 = 16 |
n1 + n2 |
2*n1 + 2*n2 = 2*(n1 + n2) |
That is, addition (and similarly, subtraction) works as is with the shifted representation.
QUIZ: Multiplication
What will be the result (using our code so far) of:
> exec "12 * 4" ghci
A. Does not compile
B. Run-time error (e.g. segmentation fault)
C. 24
D. 48
E. 96
Shifted Representation and Multiplication
We are representing a number n
by shifting it left by 1
n
has the machine representation2*n
Thus, our source values have the following _representations:
Source Value | Representation (DEC) |
---|---|
3 |
6 |
5 |
10 |
3 * 5 = 15 |
6 * 10 = 60 |
n1 * n2 |
2*n1 * 2*n2 = 4*(n1 + n2) |
Thus, multiplication ends up accumulating the factor of 2.
- Result is two times the desired one.
Strategy
Thus, our strategy for compiling arithmetic operations is:
Addition and Subtraction “just work” - as shifting “cancels out”,
Multiplication result must be “adjusted” by dividing-by-two
- i.e. right shifting by 1
Types
The source language does not change at all, for the Asm
lets add a “right shift” instruction (shr
):
data Instruction
= ...
| IShr Arg Arg
Transforms
We need only modify compileEnv
to account for the “fixing up”
compileEnv :: Env -> AnfTagE -> [Instruction]
Prim2 o v1 v2 _) = compilePrim2 env o v1 v2 compileEnv env (
where the helper compilePrim2
works for Prim2
(binary) operators and immediate arguments:
compilePrim2 :: Env -> Prim2 -> ImmE -> ImmE -> [Instruction]
Plus v1 v2 = [ IMov (Reg RAX) (immArg env v1)
compilePrim2 env IAdd (Reg RAX) (immArg env v2)
,
]Minus v1 v2 = [ IMov (Reg RAX) (immArg env v1)
compilePrim2 env ISub (Reg RAX) (immArg env v2)
,
]Times v1 v2 = [ IMov (Reg RAX) (immArg env v1)
compilePrim2 env IMul (Reg RAX) (immArg env v2)
, IShr (Reg RAX) (Const 1)
, ]
Tests
Lets take it out for a drive.
> exec' "2 * (0 - 1)"
ghci4611686018427387902
Whoa?!
Well, its easy to figure out if you look at the generated assembly:
mov rax, 4
imul rax, -2
shr rax, 1
ret
Two’s Complement
The negative result is in twos-complement format.
When we shift that right-by-one, we get the odd value
- does not “divide by two”
Decimal | Hexadecimal |
---|---|
-8 |
0xFFFFFFFFFFFFFFF8 |
2147483644 |
0x7FFFFFFFFFFFFFFC |
Solution: Signed/Arithmetic Shift
The instruction sar
shift arithmetic right does what we want, namely:
- preserves the sign-bit when shifting
- i.e. doesn’t introduce a
0
by default
Transforms Revisited
Lets add sar
to our target:
data Instruction
= ...
| ISar Arg Arg
and use it to fix the post-multiplication adjustment
- i.e. use
ISar
instead ofIShr
Times v1 v2 = [ IMov (Reg RAX) (immArg env v1)
compilePrim2 env IMul (Reg RAX) (immArg env v2)
, ISar (Reg RAX) (Const 1)
, ]
After which all is well:
> exec' "2 * (-1)"
ghci-2
3. Arithmetic Comparisons
Next, lets try to implement comparisons:
> exec "1 < 2"
ghci...
: lib/Language/Boa/Compiler.hs:(104,1)-(106,43): Non-exhaustive patterns in function compilePrim2 boa
Oops. Need to implement it first!
How to implement comparisons?
Many ways to do this:
branches
jne, jl, jg
orbit-twiddling.
Option 1: Comparisons via Branches
Key Idea:
Use the machine comparisons and branch
To implement arg1 < arg2
IF
arg1 < arg2
THEN
rax := <true>
ELSE
rax := <false>
mov rax, <arg1>
cmp rax, <arg2> # flags are set with comparison
jg false_label # if cmp-greater then false else true
mov rax, <true> # assign to RAX := true
jmp exit_label
false_label:
mov rax, <false> # assign to RAX := false
exit_label:
Option 2: Comparisons via Bit-Twiddling
Key idea:
A negative number’s most significant bit is
1
To implement arg1 < arg2
, compute arg1 - arg2
- When result is negative, MSB is
1
, ensurerax
set to0x80000001
- When result is non-negative, MSB is
0
, ensurerax
set to0x00000001
- Can extract msb by bitwise
and
with0x8000000000000000
. - Can shift msb to 32-position with
shr
- Can set tag bit by bitwise
or
with0x00000001
So compilation strategy is:
mov rax, arg1
sub rax, arg2
and rax, 0x8000000000000000 ; mask out "sign" bit (msb)
shr rax, 32 ; shift "sign" bit (msb) by 32
or rax, 0x00000001 ; set tag bit to bool
Comparisons: Implementation
Lets go and extend:
- The
Instruction
type
data Instruction
= ...
| IAnd Arg Arg
| IOr Arg Arg
- The
instrAsm
converter
instrAsm :: Instruction -> Text
IAnd a1 a2) = ...
instrAsm (IOr a1 a2) = ... instrAsm (
- The actual
compilePrim2
function
Do in class
Exercise: Comparisons via Bit-Twiddling
- Can compute
arg1 > arg2
by computingarg2 < arg1
. - Can compute
arg1 != arg2
by computingarg1 < arg2 || arg2 < arg1
- Can compute
arg1 = arg2
by computing! (arg1 != arg2)
For the above, can you figure out how to implement:
- Boolean
!
? - Boolean
||
? - Boolean
&&
?
You may find these instructions useful
4. Dynamic Checking
We’ve added support for Number
and Boolean
but we have no way to ensure that we don’t write gibberish programs like:
2 + true
or
7 < false
In fact, lets try to see what happens with our code on the above:
> exec "2 + true" ghci
Oops.
Static vs. Dynamic Type Checking
Later we will add a static type system
- that rejects meaningless programs at compile time.
Now lets add a dynamic system
- that aborts execution with wrong operands at run time.
Checking Tags at Run-Time
Here are the allowed types of operands for each primitive operation.
Operation | Op-1 | Op-2 |
---|---|---|
+ |
int |
int |
- |
int |
int |
* |
int |
int |
< |
int |
int |
> |
int |
int |
&& |
bool |
bool |
|| |
bool |
bool |
! |
bool |
|
if |
bool |
|
= |
int or bool |
int or bool |
Strategy: Asserting a Type
To check if arg
is a number
Suffices to check that the LSB is
0
If not, jump to special
error_non_int
label
For example
mov rax, arg
mov rcx, rax ; copy into rcx register
and rcx, 0x00000001 ; extract lsb
cmp rcx, 0 ; check if lsb equals 0
jne error_non_number
...
at error_non_number
we can call into a C
function:
error_non_number:
mov rdi, 0 ; pass error code
mov rsi, rax ; pass erroneous value
call error ; call run-time "error" function
Finally, the error
function is part of the run-time and looks like:
void error(long code, long v){
if (code == 0) {
"Error: expected a number but got %#010x\n", v);
fprintf(stderr,
}else if (code == 1) {
// print out message for errorcode 1 ...
}else if (code == 2) {
// print out message for errorcode 2 ...
} ...1);
exit( }
Strategy By Example
Lets implement the above in a simple file tests/output/int-check.s
section .text
extern error
extern print
global our_code_starts_here
our_code_starts_here:
mov rax, 1 ; not a valid number
mov rcx, rax ; copy into rcx register
and rcx, 0x00000001 ; extract lsb
cmp rcx, 0 ; check if lsb equals 0
jne error_non_number
error_non_number:
mov rdi, 0
mov rsi, rax
call error
Alas
make tests/output/int-check.result
... segmentation fault ...
What happened ?
Managing the Call Stack
To properly call into C functions (like error
), we must play by the rules of the C calling convention
- The local variables of an (executing) function are saved in its stack frame.
- The start of the stack frame is saved in register
rbp
, - The start of the next frame is saved in register
rsp
.
Calling Convention
We must preserve the above invariant as follows:
In the Callee
At the start of the function
push rbp ; SAVE (previous) caller's base-pointer on stack
mov rbp, rsp ; set our base-pointer using the current stack-pointer
sub rsp, 8*N ; ALLOCATE space for N local variables
At the end of the function
add rsp, 8*N0 ; FREE space for N local variables
pop rbp ; RESTORE caller's base-pointer from stack
ret ; return to caller
Fixed Strategy By Example
Lets implement the above in a simple file tests/output/int-check.s
section .text
extern error
extern print
global our_code_starts_here
our_code_starts_here:
push rbp ; save caller's base-pointer
mov rbp, rsp ; set our base-pointer
sub rsp, 1600 ; alloc '100' vars
mov rax, 1 ; not a valid number
mov rcx, rax ; copy into rcx register
and rcx, 0x00000001 ; extract lsb
cmp rcx, 0 ; check if lsb equals 0
jne error_non_number
add rsp, 1600 ; de-alloc '100' vars
pop rbp ; restore caller's base-pointer
ret
error_non_number:
mov rdi, 0
mov rsi, rax
call error
Aha, now the above works!
make tests/output/int-check.result
... expected number but got ...
Q: What NEW thing does our compiler need to compute?
Hint: Why do we sub esp, 1600
above?
Types
Lets implement the above strategy.
To do so, we need a new data type for run-time types:
data Ty = TNumber | TBoolean
a new Label
for the error
data Label
= ...
| TypeError Ty -- Type Error Labels
| Builtin Text -- Functions implemented in C
and thats it.
Transforms
The compiler must generate code to:
- Perform dynamic type checks,
- Exit by calling
error
if a failure occurs, - Manage the stack per the convention above.
1. Type Assertions
The key step in the implementation is to write a function
assertType :: Env -> IExp -> Ty -> [Instruction]
assertType env v ty= [ IMov (Reg RAX) (immArg env v)
IMov (Reg RCX) (Reg RAX)
, IAnd (Reg RCX) (HexConst 0x00000001)
, ICmp (Reg RCX) (typeTag ty)
, IJne (TypeError ty)
, ]
where typeTag
is:
typeTag :: Ty -> Arg
TNumber = HexConst 0x00000000
typeTag TBoolean = HexConst 0x00000001 typeTag
You can now splice assertType
prior to doing the actual computations, e.g.
compilePrim2 :: Env -> Prim2 -> ImmE -> ImmE -> [Instruction]
Plus v1 v2 = assertType env v1 TNumber
compilePrim2 env ++ assertType env v2 TNumber
++ [ IMov (Reg RAX) (immArg env v1)
IAdd (Reg RAX) (immArg env v2)
, ]
2. Errors
We must also add code at the TypeError TNumber
and TypeError TBoolean
labels.
errorHandler :: Ty -> Asm
=
errorHandler t ILabel (TypeError t) -- the expected-number error
[ IMov (Reg RDI) (ecode t) -- set the first "code" param,
, IMov (Reg RSI) (Reg RAX) -- set the second "value" param first,
, ICall (Builtin "error") -- call the run-time's "error" function.
,
]
ecode :: Ty -> Arg
TNumber = Const 0
ecode TBoolean = Const 1 ecode
3. Stack Management
Maintaining rsp
and rbp
We need to make sure that all our code respects the C calling convention..
To do so, just wrap the generated code, with instructions to save and restore rbp
and rsp
compileBody :: AnfTagE -> Asm
= entryCode e
compileBody e ++ compileEnv emptyEnv e
++ exitCode e
entryCode :: AnfTagE -> Asm
= [ IPush (Reg RBP) -- SAVE caller's RBP
entryCode e IMov (Reg RBP) (Reg RSP) -- SET our RBP
, ISub (Reg RSP) (Const (argBytes n)) -- ALLOC n local-vars
,
]where
= countVars e
n
exitCode :: AnfTagE -> Asm
= [ IAdd (Reg RSP) (Const (argBytes n)) -- FREE n local-vars
exitCode e IPop (Reg RBP) -- RESTORE caller's RBP
, IRet -- RETURN to caller
,
]where
= countVars e n
the rsp
needs to be a multiple of 16
so:
argBytes :: Int -> Int
= 8 * n'
argBytes n where
= if even n then n else n + 1 n'
Q: But how shall we compute countVars
?
Here’s a shady kludge:
countVars :: AnfTagE -> Int
= 100 countVars
Obviously a sleazy hack (why?), but lets use it to test everything else; then we can fix it.
4. Computing the Size of the Stack
Ok, now that everything (else) seems to work, lets work out:
countVars :: AnfTagE -> Int
Finding the exact answer is undecidable in general (CSE 105), i.e. is impossible to compute.
However, it is easy to find an overapproximate heuristic, i.e.
a value guaranteed to be larger than the than the max size,
and which is reasonable in practice.
As usual, lets see if we can work out a heuristic by example.
QUIZ
How many stack slots/vars are needed for the following program?
1 + 2
A. 0
B. 1
C. 2
QUIZ
How many stack slots/vars are needed for the following program?
let x = 1
= 2
, y = 3
, z in
+ y + z x
A. 0
B. 1
C. 2
D. 3
E. 4
QUIZ
How many stack slots/vars are needed for the following program?
if true:
let x = 1
= 2
, y = 3
, z in
+ y + z
x else:
0
A. 0
B. 1
C. 2
D. 3
E. 4
QUIZ
How many stack slots/vars are needed for the following program?
let x =
let y =
let z = 3
in z + 1
in y + 1
in x + 1
A. 0
B. 1
C. 2
D. 3
E. 4
Strategy
Let countVars e
be:
The maximum number of let-binds in scope at any point inside
e
, i.e.The maximum size of the
Env
when compilinge
Lets work it out on a case-by-case basis:
- Immediate values like
Number
orVar
- are compiled without pushing anything onto the
Env
- i.e.
countVars
= 0
- are compiled without pushing anything onto the
- Binary Operations like
Prim2 o v1 v2
take immediate values,- are compiled without pushing anything onto the
Env
- i.e.
countVars
= 0
- are compiled without pushing anything onto the
- Branches like
If v e1 e2
can go either way- can’t tell at compile-time
- i.e. worst-case is larger of
countVars e1
andcountVars e2
- Let-bindings like
Let x e1 e2
require- evaluating
e1
and - pushing the result onto the stack and then evaluating
e2
- i.e. larger of
countVars e1
and1 + countVars e2
- evaluating
Implementation
We can implement the above a simple recursive function:
countVars :: AnfTagE -> Int
If v e1 e2) = max (countVars e1) (countVars e2)
countVars (Let x e1 e2) = max (countVars e1) (1 + countVars e2)
countVars (= 0 countVars _
Naive Heuristic is Naive
The above method is quite simplistic. For example, consider the expression:
let x = 1
= 2
, y = 3
, z in
0
countVars
would tell us that we need to allocate 3
stack spaces but clearly none of the variables are actually used.
Will revisit this problem later, when looking at optimizations.
Recap
We just saw how to add support for
- Multiple datatypes (
number
andboolean
) - Calling external functions
and in doing so, learned about
- Tagged Representations
- Calling Conventions
To get some practice, in your assignment, you will add:
- Dynamic Checks for Arithmetic Overflows (see the
jo
andjno
operations) - A Primitive
print
operation implemented by a function in thec
run-time.
And next, we’ll see how to add user-defined functions.