Re: [bug-mes] MesCC assigns incorrect types to arithmetic expressions

[Jan Nieuwenhuizen]

Mark H Weaver writes:

> Currently, if I understand the code correctly, MesCC simply assigns the
> type of the first operand as the type of most (all?) binary expressions.
> For example, the type of (a + b) is the type of a, with no regard to b.
> Ditto for the relational operators.  In particular, I'm looking at
> 'ast->type' in compile.scm.

Ah, that could be; good catch.

Do you have a suggestion for a failing test (as this was something
you're aware of?).

> From an admittedly quick search, I didn't see anything about the usual
> arithmetic conversions, or the integer promotions.  Did I miss them?

I'm not sure what is usual; this is my first C compiler and I never
thought it would work so well.  As you have seen, it could do with
some refactoring; as soon as MesCC procuded a working tcc I paused
development and started on integrating mes/mescc into the guix
bootstrap...

Promotion-wise, I think that is handled in some individual cases,
like here

--8<---------------cut here---------------start------------->8---
        ((ge ,a ,b)
         (let* ((type-a (ast->type a info))
                (type-b (ast->type b info))
                (info ((binop->r info) a b 'r0-r1))
                (test->r (if (or (unsigned? type-a) (unsigned? type-b)) 'ae?->r 
'ge?->r))
                (info (append-text info (wrap-as (as info test->r))))
                (info (append-text info (wrap-as (as info 'test-r)))))
           info))
--8<---------------cut here---------------end--------------->8---

These tests (WIP branch) currently fail

--8<---------------cut here---------------start------------->8---
17-compare-unsigned-char-le
17-compare-unsigned-short-le
--8<---------------cut here---------------end--------------->8---

which may be what you're aiming at?

Thanks a lot for looking at this!

Greetings,
janneke

-- 
Jan Nieuwenhuizen <address@hidden> | GNU LilyPond http://lilypond.org
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