On Wed, Dec 04, 2019 at 05:28:04PM +0000, Ramsay Jones wrote:
> On 04/12/2019 03:06, Luc Van Oostenryck wrote:
> > +/*
> > + * check-name: autotype-ko
> > + * check-command: sparse -Wno-decl $file
> > + * checz-known-to-fail
>
> s/checz-/check-/
Hehe, that was my extra-lazy way of temporarily disabling
the tag by changing a single character ...
> > +
> > +extern char ch;
> > +extern const int ci;
> > +
> > +__auto_type i = 0; is_type(i, int);
> > +__auto_type m = 1UL; is_type(m, unsigned long);
> > +__auto_type p = &i; is_type(p, int *);
> > +__auto_type f = 0.0; is_type(f, double);
> > +__auto_type s = (struct s){0}; is_type(s, struct s);
> > +__auto_type c = ch; is_type(c, char);
> > +__auto_type ct = ci; is_type(ct, int);
>
> Hmm, this loses the 'const', ...
>
> > +__auto_type pci = &ci; is_type(pci, const int *);
>
> ... but this doesn't?
>
> Unfortunately, the gcc documentation that I have doesn't give a
> sufficiently detailed specification for me to tell if this is
> expected.
Yes, there is definitively a problem here.
In fact, I think it does the right thing because:
extern const int ci;
static void foo(void)
{
__auto_type ct = ci;
ct++;
}
rightfully complains we're trying to modify a const variable.
OTOH, the sparse-specifc type compare in the assert seems to ignore
the qualifiers at the first level. I suppose it's purposely so but ...
Another problem, maybe more important, is that the lines:
__auto_type c = ch;
__auto_type ct = ci;
at top-level are invalid because ch & ci are not constant
values/expressions and no warnings are issued.
It's orthogonal to __auto_type but quite annoying.
Thanks for noticing all this!
-- Luc
