TC-6 Canonicalization Samples
The first task in TC-6 is getting rid of all the
eseq. To do this, you have to move the statement part of an
eseq at the end of the current sequence point, and keeping
the expression part in place.
Compare for instance the HIR to the LIR in the following case.
let
function print_ints(a: int, b: int) =
(print_int(a); print(", "); print_int(b); print("\n"))
var a := 0
in
print_ints(1, (a := a + 1; a))
end
One possible HIR translation is:
$ tc -eH preincr-1.tig
/* == High Level Intermediate representation. == */
label l1
", "
label l2
"\n"
# Routine: print_ints
label l0
# Prologue
move
temp t2
temp fp
move
temp fp
temp sp
move
temp sp
binop sub
temp sp
const 4
move
mem
temp fp
temp i0
move
temp t0
temp i1
move
temp t1
temp i2
# Body
seq
sxp
call
name print_int
temp t0
call end
sxp
call
name print
name l1
call end
sxp
call
name print_int
temp t1
call end
sxp
call
name print
name l2
call end
seq end
# Epilogue
move
temp sp
temp fp
move
temp fp
temp t2
label end
# Routine: _main
label main
# Prologue
# Body
seq
seq
move
temp t3
const 0
sxp
call
name l0
temp fp
const 1
eseq
move
temp t3
binop add
temp t3
const 1
temp t3
call end
seq end
sxp
const 0
seq end
# Epilogue
label end
$ echo $?
0
A possible canonicalization is then:
$ tc -eL preincr-1.tig
/* == Low Level Intermediate representation. == */
label l1
", "
label l2
"\n"
# Routine: print_ints
label l0
# Prologue
move
temp t2
temp fp
move
temp fp
temp sp
move
temp sp
binop sub
temp sp
const 4
move
mem
temp fp
temp i0
move
temp t0
temp i1
move
temp t1
temp i2
# Body
seq
label l3
sxp
call
name print_int
temp t0
call end
sxp
call
name print
name l1
call end
sxp
call
name print_int
temp t1
call end
sxp
call
name print
name l2
call end
label l4
seq end
# Epilogue
move
temp sp
temp fp
move
temp fp
temp t2
label end
# Routine: _main
label main
# Prologue
# Body
seq
label l5
move
temp t3
const 0
move
temp t5
temp fp
move
temp t3
binop add
temp t3
const 1
sxp
call
name l0
temp t5
const 1
temp t3
call end
label l6
seq end
# Epilogue
label end
$ echo $?
0
The example above is simple because 1 commutes with
(a := a + 1; a): the order does not matter. But if you change
the 1 into a, then you cannot exchange a and
(a := a + 1; a), so the translation is different. Compare the
previous LIR with the following.
let
function print_ints(a: int, b: int) =
(print_int(a); print(", "); print_int(b); print("\n"))
var a := 0
in
print_ints(a, (a := a + 1; a))
end
$ tc -eL preincr-2.tig
/* == Low Level Intermediate representation. == */
label l1
", "
label l2
"\n"
# Routine: print_ints
label l0
# Prologue
move
temp t2
temp fp
move
temp fp
temp sp
move
temp sp
binop sub
temp sp
const 4
move
mem
temp fp
temp i0
move
temp t0
temp i1
move
temp t1
temp i2
# Body
seq
label l3
sxp
call
name print_int
temp t0
call end
sxp
call
name print
name l1
call end
sxp
call
name print_int
temp t1
call end
sxp
call
name print
name l2
call end
label l4
seq end
# Epilogue
move
temp sp
temp fp
move
temp fp
temp t2
label end
# Routine: _main
label main
# Prologue
# Body
seq
label l5
move
temp t3
const 0
move
temp t6
temp fp
move
temp t5
temp t3
move
temp t3
binop add
temp t3
const 1
sxp
call
name l0
temp t6
temp t5
temp t3
call end
label l6
seq end
# Epilogue
label end
$ echo $?
0
As you can see, the output is the same for the HIR and the LIR.
$ tc -eH preincr-2.tig > preincr-2.hir
$ echo $?
0
$ havm preincr-2.hir
0, 1
$ echo $?
0
$ tc -eL preincr-2.tig > preincr-2.lir
$ echo $?
0
$ havm preincr-2.lir
0, 1
$ echo $?
0
Be very careful when dealing with mem. For instance, rewriting
something like:
call(foo, eseq(move(temp t, const 51), temp t))
into
move temp t1, temp t
move temp t, const 51
call(foo, temp t)
is wrong: temp t is not a subexpression, rather it is being
defined here. You should produce:
move temp t, const 51
call(foo, temp t)
Another danger is the handling of move(mem, ). For instance:
move(mem foo, x)
must be rewritten into:
move(temp t, foo)
move(mem(temp t), x)
not as:
move(temp t, mem(foo))
move(temp t, x)
In other words, the first subexpression of move(mem(foo), ) is
foo, not mem(foo). The following example is a good crash
test against this problem.
let type int_array = array of int
var tab := int_array [2] of 51
in
tab[0] := 100;
tab[1] := 200;
print_int(tab[0]); print("\n");
print_int(tab[1]); print("\n")
end
$ tc -eL move-mem.tig > move-mem.lir
$ echo $?
0
$ havm move-mem.lir
100
200
$ echo $?
0
You also ought to get rid of nested calls.
print(chr(ord("\n")))
$ tc -L nested-calls.tig
/* == Low Level Intermediate representation. == */
label l0
"\n"
# Routine: _main
label main
# Prologue
# Body
seq
label l1
move
temp t1
call
name ord
name l0
call end
move
temp t2
call
name chr
temp t1
call end
sxp
call
name print
temp t2
call end
label l2
seq end
# Epilogue
label end
$ echo $?
0
There are only two valid call forms: sxp(call(...)), and
move(temp(...), call(...)).
Contrary to C, the HIR and LIR always denote the same value. For instance the following Tiger code:
let
var a := 1
function a(t: int) : int =
(a := a + 1;
print_int(t); print(" -> "); print_int(a); print("\n");
a)
var b := a(1) + a(2) * a(3)
in
print_int(b); print("\n")
end
should always produce:
$ tc -L seq-point.tig > seq-point.lir
$ echo $?
0
$ havm seq-point.lir
1 -> 2
2 -> 3
3 -> 4
14
$ echo $?
0
independently of which IR you ran. It has nothing to do with operator precedence!
In C, you have no such guarantee: the following program can give different results with different compilers and/or on different architectures.
#include <stdio.h>
int a_ = 1;
int
a(int t)
{
++a_;
printf("%d -> %d\n", t, a_);
return a_;
}
int
main(void)
{
int b = a(1) + a(2) * a(3);
printf("%d\n", b);
return 0;
}