1

kconfig: remove 'e1' and 'e2' macros from expression deduplication

I do not think the macros 'e1' and 'e2' are readable.

The statement:

    e1 = expr_alloc_symbol(...);

affects the caller's variable, but this is not sufficiently clear from the code.

Remove the macros. No functional change intended.

Signed-off-by: Masahiro Yamada <masahiroy@kernel.org>
This commit is contained in:
Masahiro Yamada 2024-07-08 00:38:07 +09:00
parent 94a4b0a4cb
commit 3c2f84ceda

View File

@ -135,9 +135,6 @@ void expr_free(struct expr *e)
static int trans_count;
#define e1 (*ep1)
#define e2 (*ep2)
/*
* expr_eliminate_eq() helper.
*
@ -150,38 +147,38 @@ static void __expr_eliminate_eq(enum expr_type type, struct expr **ep1, struct e
{
/* Recurse down to leaves */
if (e1->type == type) {
__expr_eliminate_eq(type, &e1->left.expr, &e2);
__expr_eliminate_eq(type, &e1->right.expr, &e2);
if ((*ep1)->type == type) {
__expr_eliminate_eq(type, &(*ep1)->left.expr, ep2);
__expr_eliminate_eq(type, &(*ep1)->right.expr, ep2);
return;
}
if (e2->type == type) {
__expr_eliminate_eq(type, &e1, &e2->left.expr);
__expr_eliminate_eq(type, &e1, &e2->right.expr);
if ((*ep2)->type == type) {
__expr_eliminate_eq(type, ep1, &(*ep2)->left.expr);
__expr_eliminate_eq(type, ep1, &(*ep2)->right.expr);
return;
}
/* e1 and e2 are leaves. Compare them. */
/* *ep1 and *ep2 are leaves. Compare them. */
if (e1->type == E_SYMBOL && e2->type == E_SYMBOL &&
e1->left.sym == e2->left.sym &&
(e1->left.sym == &symbol_yes || e1->left.sym == &symbol_no))
if ((*ep1)->type == E_SYMBOL && (*ep2)->type == E_SYMBOL &&
(*ep1)->left.sym == (*ep2)->left.sym &&
((*ep1)->left.sym == &symbol_yes || (*ep1)->left.sym == &symbol_no))
return;
if (!expr_eq(e1, e2))
if (!expr_eq(*ep1, *ep2))
return;
/* e1 and e2 are equal leaves. Prepare them for elimination. */
/* *ep1 and *ep2 are equal leaves. Prepare them for elimination. */
trans_count++;
expr_free(e1); expr_free(e2);
expr_free(*ep1); expr_free(*ep2);
switch (type) {
case E_OR:
e1 = expr_alloc_symbol(&symbol_no);
e2 = expr_alloc_symbol(&symbol_no);
*ep1 = expr_alloc_symbol(&symbol_no);
*ep2 = expr_alloc_symbol(&symbol_no);
break;
case E_AND:
e1 = expr_alloc_symbol(&symbol_yes);
e2 = expr_alloc_symbol(&symbol_yes);
*ep1 = expr_alloc_symbol(&symbol_yes);
*ep2 = expr_alloc_symbol(&symbol_yes);
break;
default:
;
@ -219,29 +216,26 @@ static void __expr_eliminate_eq(enum expr_type type, struct expr **ep1, struct e
*/
void expr_eliminate_eq(struct expr **ep1, struct expr **ep2)
{
if (!e1 || !e2)
if (!*ep1 || !*ep2)
return;
switch (e1->type) {
switch ((*ep1)->type) {
case E_OR:
case E_AND:
__expr_eliminate_eq(e1->type, ep1, ep2);
__expr_eliminate_eq((*ep1)->type, ep1, ep2);
default:
;
}
if (e1->type != e2->type) switch (e2->type) {
if ((*ep1)->type != (*ep2)->type) switch ((*ep2)->type) {
case E_OR:
case E_AND:
__expr_eliminate_eq(e2->type, ep1, ep2);
__expr_eliminate_eq((*ep2)->type, ep1, ep2);
default:
;
}
e1 = expr_eliminate_yn(e1);
e2 = expr_eliminate_yn(e2);
*ep1 = expr_eliminate_yn(*ep1);
*ep2 = expr_eliminate_yn(*ep2);
}
#undef e1
#undef e2
/*
* Returns true if 'e1' and 'e2' are equal, after minor simplification. Two
* &&/|| expressions are considered equal if every operand in one expression
@ -564,59 +558,55 @@ static struct expr *expr_join_and(struct expr *e1, struct expr *e2)
*/
static void expr_eliminate_dups1(enum expr_type type, struct expr **ep1, struct expr **ep2)
{
#define e1 (*ep1)
#define e2 (*ep2)
struct expr *tmp;
/* Recurse down to leaves */
if (e1->type == type) {
expr_eliminate_dups1(type, &e1->left.expr, &e2);
expr_eliminate_dups1(type, &e1->right.expr, &e2);
if ((*ep1)->type == type) {
expr_eliminate_dups1(type, &(*ep1)->left.expr, ep2);
expr_eliminate_dups1(type, &(*ep1)->right.expr, ep2);
return;
}
if (e2->type == type) {
expr_eliminate_dups1(type, &e1, &e2->left.expr);
expr_eliminate_dups1(type, &e1, &e2->right.expr);
if ((*ep2)->type == type) {
expr_eliminate_dups1(type, ep1, &(*ep2)->left.expr);
expr_eliminate_dups1(type, ep1, &(*ep2)->right.expr);
return;
}
/* e1 and e2 are leaves. Compare and process them. */
/* *ep1 and *ep2 are leaves. Compare and process them. */
if (e1 == e2)
if (*ep1 == *ep2)
return;
switch (e1->type) {
switch ((*ep1)->type) {
case E_OR: case E_AND:
expr_eliminate_dups1(e1->type, &e1, &e1);
expr_eliminate_dups1((*ep1)->type, ep1, ep1);
default:
;
}
switch (type) {
case E_OR:
tmp = expr_join_or(e1, e2);
tmp = expr_join_or(*ep1, *ep2);
if (tmp) {
expr_free(e1); expr_free(e2);
e1 = expr_alloc_symbol(&symbol_no);
e2 = tmp;
expr_free(*ep1); expr_free(*ep2);
*ep1 = expr_alloc_symbol(&symbol_no);
*ep2 = tmp;
trans_count++;
}
break;
case E_AND:
tmp = expr_join_and(e1, e2);
tmp = expr_join_and(*ep1, *ep2);
if (tmp) {
expr_free(e1); expr_free(e2);
e1 = expr_alloc_symbol(&symbol_yes);
e2 = tmp;
expr_free(*ep1); expr_free(*ep2);
*ep1 = expr_alloc_symbol(&symbol_yes);
*ep2 = tmp;
trans_count++;
}
break;
default:
;
}
#undef e1
#undef e2
}
/*