[9fans] fidtab in u9fs

[9fans] fidtab in u9fs

[Andrew Simmons]

I've been looking at the source for u9fs, and am a bit puzzled by the
table of fids and the newfid function. The table is declared as

Fidtab *fidtab[1];

and the business part of newfid is:

f = emalloc(sizeof(*f));
f->next = fidtab[fid%nelem(fidtab)];
if(f->next)
    f->next->prev = f;
fidtab[fid%nelem(fidtab)] = f;
f->fid = fid;

As far as I can see, nelem(fidtab) is 1, so that fidtab is always
indexed by 0, and there is only ever a single element in the table,
whose next and prev elements point at itself. And if so, what's the
point of the loop in lookupfid?

I'm probably being really thick here, but what am I missing?

Re: [9fans] fidtab in u9fs

[Russ Cox]

There can be more than one element in the list --
if you call newfid twice then there will be two fids
in the list.  You should look at removefid too.

It would be a more effective hash table if fidtab
were declared with more than one element.
That was probably debugging that never got taken out.

Russ

Re: [9fans] fidtab in u9fs

[Andrew Simmons]

> It would be a more effective hash table if fidtab
> were declared with more than one element.

Thanks. I was fixating on the declaration so much that I didn't read
the code properly. Mea maxima culpa.