On May 13, 2008, at 12:57 AM, john skaller wrote:
> The likely problem is that no two languages have the same architecture,
> if they did .. why bother with another language?
>
> C and C++ are exceptions here, since one is derived from another.
Yeah, I understand that. Got a better idea? We might be able to go
directly to source, but then we'd have to handle the SSA stuff
ourselves (do we do this already?). There's also the overhead of
having to write out the assembly then load it back in for llvm to
process it. So, to walk through a basic example. Say we have this
simple code:
fun foo (a:double, b:double) => a*a + 2.0*a*b + b*b;
val a = 1.0;
val b = 2.0;
val c = (foo (a, b)) + 3.0;
printd d;
If we do a straightforward translation to llvm we should result in
approximately this:
define double @foo(double %a, double %b) {
entry:
%tmp = mul double %a, %a
%tmp1 = mul double %a, 2.0
%tmp2 = mul double %tmp1, %b
%tmp3 = add double %tmp, %tmp2
%tmp4 = mul double %b, %b
%tmp5 = add double %tmp3, %tmp4
ret double %tmp5
}
declare double @printd(double)
define i32 @main(i32 %argc, i8 **argv) {
entry:
%a = 1.0
%b = 2.0
%tmp = call double @foo(double %a, double %b)
%c = add %tmp, 3.0
call void @printd(double %c)
ret i32 0
}
For this simple block of code, I think we could probably implement the
standard library like this:
type double = "double";
fun add (a:double, b:double) = "add ?1 %$1, %$2";
fun mul (a:double, b:double) = "mul ?1 %$1, %$2";
proc printd: double -> unit = "call void @printd(?1 $1)";
However, will it work with control flow? If we then have:
fun foo: unit -> double = "...";
fun bar: unit -> double = "...";
fun baz: unit -> double = "...";
fun boo (a:double) = {
val b = if a == 0 then foo() else bar();
return b + boo();
}
We need to generate this:
declare double @foo()
declare double @bar()
declare double @baz()
define double @boo(double %a) {
entry:
%tmp = fcmp one %a, 0
br il %tmp, label %then, label %else
then:
%tmp1 = call double foo()
br label %after
else:
%tmp2 = call double bar()
br label %after
after:
%tmp3 = phi double [ %tmp1, %then ], [ %tmp2, %else ]
%tmp4 = call double baz()
%tmp5 = add double %tmp3, %tmp4
ret double %tmp5
}
Hmm. That might work as well. I'm not sure if it'd capture all the
semantics that we need, but it might be a good way to start.
In the long run, we still could go with the scheme-based FFI, but just
make no attempt to share the same variants as the c++ FFI.
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