kconfig: Document important expression functions

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@gmail.com>
Signed-off-by: Masahiro Yamada <yamada.masahiro@socionext.com>

1 files changed, 106 insertions, 0 deletions

diff --git a/scripts/kconfig/expr.c b/scripts/kconfig/expr.c
index ed29bad1f03a..fd8a416ceab7 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;