diff options
Diffstat (limited to 'scripts/kconfig/expr.c')
| -rw-r--r-- | scripts/kconfig/expr.c | 106 |
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; |