On Tue, Jun 18, 2024 at 08:25:34AM -0700, Junio C Hamano wrote:
> shejialuo <shejialuo@xxxxxxxxx> writes:
>
> > On Tue, Jun 18, 2024 at 04:38:07AM -0400, Karthik Nayak wrote:
> >> shejialuo <shejialuo@xxxxxxxxx> writes:
> >>
> >> [snip]
> >>
> >> > struct fsck_options {
> >> > + /*
> >> > + * Reorder the fields to allow `fsck_ref_options` to use
> >> > + * the interfaces using `struct fsck_options`.
> >> > + */
> >>
> >> Why is this added? It makes sense to have it in the commit message
> >> because it talks about the change, but why make it persistent in the
> >> code?
> >>
> >
> > I explicitly add this comments due to the following reasons:
> >
> > 1. If someone needs to change the `fsck_options`, without this comment,
> > he might be just add some new fields at top. Although the change will
> > fail the tests here, I think we should mention this in code.
>
> Do you mean you plan to take advantage of the fact that early
> members of two structures are the same? IOW, if there is a function
> that takes a pointer to smaller fsck_refs_options, you plan to pass
> a pointer to fsck_options from some callers, e.g.
>
> extern void func(struct fsck_refs_options *);
> void a_caller(struct fsck_options *o)
> {
> func((struct fsck_options *)o);
> ...
>
> If that is the case, then ...
>
> Do not do that.
>
> Your data structure design is broken. Instead you would do this:
>
> struct fsck_options {
> struct fsck_refs_options refs;
> ... other members ...
> };
> void a_caller(struct fsck_options *o)
> {
> func(&o->refs);
> ...
Well, I totally agree with this. It's bad to convert the pointer type. I
will change the code in this patch to make it OK.
Thanks for your advice.
