On 01/09/2025 20:31, Kees Cook wrote:
On Mon, Sep 01, 2025 at 08:20:18PM +0200, Vegard Nossum wrote:
On 01/09/2025 18:56, Kees Cook wrote:
@@ -459,13 +462,15 @@ void sym_calc_value(struct symbol *sym)
sym_calc_choice(choice_menu);
newval.tri = sym->curr.tri;
} else {
- if (sym->visible != no) {
+ if (sym->usable) {
/* if the symbol is visible use the user value
* if available, otherwise try the default value
*/
if (sym_has_value(sym)) {
+ tristate value = sym->transitional ?
+ sym->def[S_DEF_USER].tri : sym->visible;
newval.tri = EXPR_AND(sym->def[S_DEF_USER].tri,
- sym->visible);
+ value);
This looks a bit odd to me. Just thinking out loud: your new logic is
there to be able to use a value even though it's not visible. In the
case where it's transitional you use the .config value instead of the
condition that makes it visible.
Could you simply change sym_calc_visibility() instead to always return
'yes' when the symbol is transitional? Wouldn't that simplify everything
in sym_calc_value()?
It's a tristate, so "m" is also possible besides "y". (sym->visible is
also a tristate. 🙂
That would be fine, right?
We'd pass the if (sym->visible != no) check... we'd do the
newval.tri = EXPR_AND(sym->def[S_DEF_USER].tri, sym->visible);
EXPR_AND() is basically min() (with n=0, m=1, y=2), so effectively it
would end up doing
newval.tri = min(sym->def[S_DEF_USER].tri, 2);
which is the same as
newval.tri = sym->def[S_DEF_USER].tri;
That's what your code is currently doing too, but in a much more
roundabout way.
Right, it was this:
newval.tri = EXPR_AND(sym->def[S_DEF_USER].tri, sym->visible);
But I made it effectively:
if (sym->transitional)
newval.tri = EXPR_AND(sym->def[S_DEF_USER].tri, sym->def[S_DEF_USER].tri);
else
newval.tri = EXPR_AND(sym->def[S_DEF_USER].tri, sym->visible);
That first "if" is kind of pointless. I just sent the v3 before I saw
this email. :P
I was trying to avoid yet more indentation, but I could change it to:
if (sym->transitional)
newval.tri = sym->def[S_DEF_USER].tri;
else
newval.tri = EXPR_AND(sym->def[S_DEF_USER].tri,
sym->visible);
?
If you change sym_calc_visibility() to always return 'yes' for
transitional values then I don't think you need to touch
sym_calc_value() at all.
Vegard
