Many of these functions are quite the head scratchers if you don't know
what they're trying to do. Document them.
Also make it clear which functions rewrite expressions in-place and
which return new expressions. This prevents memory errors.
No functional changes. Only comments added.
Signed-off-by: Ulf Magnusson <ulfalizer@xxxxxxxxx>
---
scripts/kconfig/expr.c | 106 +++++++++++++++++++++++++++++++++++++++++++++++++
1 file changed, 106 insertions(+)
diff --git a/scripts/kconfig/expr.c b/scripts/kconfig/expr.c
index cbf4996..5965ecd 100644
--- a/scripts/kconfig/expr.c
+++ b/scripts/kconfig/expr.c
@@ -138,8 +138,18 @@ static int trans_count;
#define e1 (*ep1)
#define e2 (*ep2)
+/*
+ * expr_eliminate_eq() helper.
+ *
+ * Walks the two expression trees given in 'ep1' and 'ep2'. Any node that does
+ * not have type 'type' (E_OR/E_AND) is considered a leaf, and is compared
+ * against all other leaves. Two equal leaves are both replaced with either 'y'
+ * or 'n' as appropriate for 'type', to be eliminated later.
+ */
static void __expr_eliminate_eq(enum expr_type type, struct expr **ep1, struct expr **ep2)
{
+ /* Recurse down to leaves */
+
if (e1->type == type) {
__expr_eliminate_eq(type, &e1->left.expr, &e2);
__expr_eliminate_eq(type, &e1->right.expr, &e2);
@@ -150,12 +160,18 @@ static void __expr_eliminate_eq(enum expr_type type, struct expr **ep1, struct e
__expr_eliminate_eq(type, &e1, &e2->right.expr);
return;
}
+
+ /* e1 and e2 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))
return;
if (!expr_eq(e1, e2))
return;
+
+ /* e1 and e2 are equal leaves. Prepare them for elimination. */
+
trans_count++;
expr_free(e1); expr_free(e2);
switch (type) {
@@ -172,6 +188,35 @@ static void __expr_eliminate_eq(enum expr_type type, struct expr **ep1, struct e
}
}
+/*
+ * Rewrites the expressions 'ep1' and 'ep2' to remove operands common to both.
+ * Example reductions:
+ *
+ * ep1: A && B -> ep1: y
+ * ep2: A && B && C -> ep2: C
+ *
+ * ep1: A || B -> ep1: n
+ * ep2: A || B || C -> ep2: C
+ *
+ * ep1: A && (B && FOO) -> ep1: FOO
+ * ep2: (BAR && B) && A -> ep2: BAR
+ *
+ * ep1: A && (B || C) -> ep1: y
+ * ep2: (C || B) && A -> ep2: y
+ *
+ * Comparisons are done between all operands at the same "level" of && or ||.
+ * For example, in the expression 'e1 && (e2 || e3) && (e4 || e5)', the
+ * following operands will be compared:
+ *
+ * - 'e1', 'e2 || e3', and 'e4 || e5', against each other
+ * - e2 against e3
+ * - e4 against e5
+ *
+ * Parentheses are irrelevant within a single level. 'e1 && (e2 && e3)' and
+ * '(e1 && e2) && e3' are both a single level.
+ *
+ * See __expr_eliminate_eq() as well.
+ */
void expr_eliminate_eq(struct expr **ep1, struct expr **ep2)
{
if (!e1 || !e2)
@@ -197,6 +242,12 @@ void expr_eliminate_eq(struct expr **ep1, struct expr **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
+ * equals some operand in the other (operands do not need to appear in the same
+ * order), recursively.
+ */
static int expr_eq(struct expr *e1, struct expr *e2)
{
int res, old_count;
@@ -243,6 +294,17 @@ static int expr_eq(struct expr *e1, struct expr *e2)
return 0;
}
+/*
+ * Recursively performs the following simplifications in-place (as well as the
+ * corresponding simplifications with swapped operands):
+ *
+ * expr && n -> n
+ * expr && y -> expr
+ * expr || n -> expr
+ * expr || y -> y
+ *
+ * Returns the optimized expression.
+ */
static struct expr *expr_eliminate_yn(struct expr *e)
{
struct expr *tmp;
@@ -516,12 +578,21 @@ static struct expr *expr_join_and(struct expr *e1, struct expr *e2)
return NULL;
}
+/*
+ * expr_eliminate_dups() helper.
+ *
+ * Walks the two expression trees given in 'ep1' and 'ep2'. Any node that does
+ * not have type 'type' (E_OR/E_AND) is considered a leaf, and is compared
+ * against all other leaves to look for simplifications.
+ */
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);
@@ -532,6 +603,9 @@ static void expr_eliminate_dups1(enum expr_type type, struct expr **ep1, struct
expr_eliminate_dups1(type, &e1, &e2->right.expr);
return;
}
+
+ /* e1 and e2 are leaves. Compare and process them. */
+
if (e1 == e2)
return;
@@ -568,6 +642,17 @@ static void expr_eliminate_dups1(enum expr_type type, struct expr **ep1, struct
#undef e2
}
+/*
+ * Rewrites 'e' in-place to remove ("join") duplicate and other redundant
+ * operands.
+ *
+ * Example simplifications:
+ *
+ * A || B || A -> A || B
+ * A && B && A=y -> A=y && B
+ *
+ * Returns the deduplicated expression.
+ */
struct expr *expr_eliminate_dups(struct expr *e)
{
int oldcount;
@@ -584,6 +669,7 @@ struct expr *expr_eliminate_dups(struct expr *e)
;
}
if (!trans_count)
+ /* No simplifications done in this pass. We're done */
break;
e = expr_eliminate_yn(e);
}
@@ -591,6 +677,12 @@ struct expr *expr_eliminate_dups(struct expr *e)
return e;
}
+/*
+ * Performs various simplifications involving logical operators and
+ * comparisons.
+ *
+ * Allocates and returns a new expression.
+ */
struct expr *expr_transform(struct expr *e)
{
struct expr *tmp;
@@ -805,6 +897,20 @@ bool expr_depends_symbol(struct expr *dep, struct symbol *sym)
return false;
}
+/*
+ * Inserts explicit comparisons of type 'type' to symbol 'sym' into the
+ * expression 'e'.
+ *
+ * Examples transformations for type == E_UNEQUAL, sym == &symbol_no:
+ *
+ * A -> A!=n
+ * !A -> A=n
+ * A && B -> !(A=n || B=n)
+ * A || B -> !(A=n && B=n)
+ * A && (B || C) -> !(A=n || (B=n && C=n))
+ *
+ * Allocates and returns a new expression.
+ */
struct expr *expr_trans_compare(struct expr *e, enum expr_type type, struct symbol *sym)
{
struct expr *e1, *e2;
--
2.7.4
--
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