void count_cases(SNODE *stmt, struct cases *cs)
{
while (stmt)
{
switch(stmt->stype) {
case st_tryblock:
break;
case st_throw:
break;
case st_return:
case st_expr:
break;
case st_while:
case st_do:
count_cases(stmt->s1,cs);
break;
case st_for:
count_cases(stmt->s1,cs);
break;
case st_if:
count_cases(stmt->s1,cs);
count_cases(stmt->s2,cs);
break;
case st_switch:
break;
case st_block:
count_cases(stmt->exp,cs);
break;
case st_asm:
break;
case st_case:
if (!stmt->s2)
{
cs->count++;
if (stmt->switchid < cs->bottom)
cs->bottom = stmt->switchid;
if (stmt->switchid > cs->top)
cs->top = stmt->switchid;
}
break;
}
stmt = stmt->next;
}
}
/* Allocate the structures for handling a sequence of expressions
* to be combined with AND. Expand the structure into a series of
* cases. Do not start the computations yet, but prepare them.
* The sequence of expressions is considered a NULL-terminated
* array of NUL-terminated strings.
*
* This routine returns NULL in case of error, and will log the
* expression that caused the failure. In case of success, an
* expanded expression is returned, that should be cleaned up
* with free() as soon as it is ready.
*/
static struct valexp *allocate_valexp (char **and_expressions) {
int allcases = 1;
char **andexp;
struct valexp *retval = NULL;
int memsz;
//
// Count the individual expressions' sizes, and compute the total
for (andexp=and_expressions; (*andexp) != NULL; andexp++) {
int explen = strlen (*andexp);
int parsed;
int newcases = count_cases (*andexp, explen, 0, &parsed);
if ((newcases == -1) || (parsed != explen)) {
//TODO// Expression will not be the same anymore
tlog (TLOG_USER, LOG_NOTICE, "Syntax error in logic expression, treat as False: %s", *andexp);
return NULL;
}
allcases = allcases * newcases;
}
//
// Allocate memory for the overal expression
// Since all cases are initialised to all-zeroes, they represent
// TRUE in all these cases. Expansion should add restrictions.
memsz = sizeof (struct valexp) + sizeof (struct valexp_case) * (allcases - 1);
retval = (struct valexp *) malloc (memsz);
if (retval == NULL) {
//TODO// Expression will not be the same anymore
tlog (TLOG_TLS, LOG_NOTICE, "Out of memory expanding logic expressions");
return NULL;
}
memset (retval, 0, memsz);
retval->numcases = allcases;
retval->numcases_incomplete = allcases;
//
// Expand the expressions to the form (OR (AND ... ~...) (AND ...) ...)
// This relies on:
// - De Morgan to fold inversion inward
// - Distribution laws for AND / OR
// - Rewriting of ? to a combination and AND / OR
// - No cases as the zero element of OR to represent FALSE
// - A complete case to represent TRUE
// Note that the and_expressions are pushed into each' OR caselist
// Note that all cases are initialised AND-ready (they're all TRUE)
for (andexp=and_expressions; *andexp != NULL; andexp++) {
//TODO// EXPAND_EXPRESSION_INTO_EXISTING_WITH_AND
}
//
// Return successfully.
return retval;
}
void genxswitch(STATEMENT *stmt, SYMBOL *funcsp)
/*
* analyze and generate best switch statement.
*/
{
int oldbreak, i;
struct cases cs;
IMODE *ap, *ap3;
#ifdef USE_LONGLONG
ULLONG_TYPE a = 1;
#endif
oldbreak = breaklab;
breaklab = stmt->breaklabel;
memset(&cs,0,sizeof(cs));
#ifndef USE_LONGLONG
cs.top = INT_MIN;
cs.bottom = INT_MAX;
#else
cs.top = (a << 63); /* LLONG_MIN*/
cs.bottom = cs.top - 1; /* LLONG_MAX*/
#endif
count_cases(stmt->cases,&cs) ;
cs.top++;
ap3 = gen_expr(funcsp, stmt->select, F_VOL | F_NOVALUE, ISZ_UINT);
ap = LookupLoadTemp(NULL, ap3);
if (ap != ap3)
{
IMODE *barrier;
if (stmt->select->isatomic)
{
barrier = doatomicFence(funcsp, stmt->select, NULL);
}
gen_icode(i_assn, ap, ap3, NULL);
if (stmt->select->isatomic)
{
doatomicFence(funcsp, stmt->select, barrier);
}
}
gen_icode2(i_coswitch, make_immed(ISZ_NONE,cs.count), ap, make_immed(ISZ_NONE,cs.top - cs.bottom), stmt->label);
gather_cases(stmt->cases,&cs);
qsort(cs.ptrs, cs.count, sizeof(cs.ptrs[0]), gcs_compare);
for (i = 0; i < cs.count; i++)
{
gen_icode2(i_swbranch,0,make_immed(ISZ_NONE,cs.ptrs[i].id),0,cs.ptrs[i].label);
}
breaklab = oldbreak;
}
int analyzeswitch(SNODE *stmt, struct cases *cs)
{
int size = natural_size(stmt->exp);
count_cases(stmt->s1,cs) ;
cs->top++;
if (cs->count == 0)
return (0);
if (cs->count < 5)
return 3;
if (cs->count *10 / (cs->top - cs->bottom) >= 8)
return (1);
// do a simple switch instead of a binary if it is a long long
if (size == BESZ_QWORD || size == - BESZ_QWORD)
return 3;
return (2);
}
/* Count the number of cases in a validation expression; that is, after
* folding all the | and ? to the outside for fastest-possible evaluation
* caused by the shortest computation path to a positive outcome. We
* care less about refusals -- they may take time. That'll teach them :)
*
* This routine returns -1 if the syntax or anything else in the
* validation expression is not in order; otherwise it returns a
* case count >= 0. As a few (not so) special cases, the logic value
* FALSE written as 0 or 1~ is returned as zero cases, because FALSE is
* the zero element for logical OR; and the logic value TRUE, which
* is written as 1 or 0~, is returned as one case that will be
* setup during expression expansion to have succeeded immediately,
* namely as a case with nothing left to be computed, and with no
* negative or positive actions that were not satisfied. Since TRUE
* is the zero element for logical AND, it is often used as a default
* in AND compositions, and will then combine with further cases when
* it the validation expressions are being expanded. It makes sense
* in algebra, and so it makes sense in real life.
*
* While counting, some useful information is collected and stored by
* overwriting binary operators with a byte that has the high bit set
* to form a recognisable OPC_xxx code. There are flags OPF_xxx to
* indicate trivia found while counting, namely whether left/right
* branches have zero cases, and whether zero cases trickle up. This
* information is of great use for the interleaving algorithm. This
* also means that count_cases() needs to run before expand_cases_rec().
*
* Even though the tree is modified, count_cases() can still pass
* through it, provided that it did not return -1 before. This means
* that it should not be made to work on constant strings, but it can
* be made to work on the same (global/static) variables repeatedly or
* even concurrently; it is designed to be idempotent and re-entrant.
*
* On success, the function fills the parsed value with the number
* of characters that were parsed (in a range of 0 up to vallen)
* starting from the end of the expression -- since it is reverse
* Polish notation. The top-call to this function should return in
* the output parameter parsed the same value as the vallen it was
* called with. Otherwise, the validation expression did not accept
* the entire string passed to it as valexp / vallen.
*/
static int count_cases (char *valexpstr, int vallen, int invert, int *parsed) {
int case0p, case0n, case1, case2;
int pars0, pars1, pars2;
uint8_t *opcp;
uint8_t opc;
uint8_t zero_1st = invert? OPF_ONE_FIRST: OPF_ZERO_FIRST;
uint8_t zero_2nd = invert? OPF_ONE_SECOND: OPF_ZERO_SECOND;
int retval = -1;
//
// Ensure that the validation expression is non-empty
if (vallen <= 0) {
return -1;
}
vallen--;
opcp = &valexpstr [vallen];
opc = *opcp;
switch (opc) {
case '&':
opc = OPC_AND;
break;
case '|':
opc = OPC_OR;
break;
case '?':
opc = OPC_IF;
break;
default:
if (opc & 0x80) {
opc = opc & OPC_MASK;
}
break;
}
//
// Find one of | or & or ~ or ?
switch (opc) {
case OPC_OR:
case OPC_AND:
case1 = count_cases (valexpstr, vallen, invert, &pars1);
vallen -= pars1;
case2 = count_cases (valexpstr, vallen, invert, &pars2);
*parsed = 1 + pars1 + pars2;
if ((case1 == -1) || (case2 == -1)) {
*opcp = ERR_BINOP;
return -1;
} else if (opc == (invert? OPC_AND: OPC_OR)) {
retval = case1 + case2;
} else {
retval = case1 * case2;
}
*opcp = opc;
if (case1 == 0) {
*opcp |= zero_1st;
}
if (case2 == 0) {
*opcp |= zero_2nd;
}
return retval;
case OPC_IF:
// Count the test case twice; once positive and once negative
// "eti?" reads as "IF i THEN t ELSE e" or like in C, "i?t:e"
case0p = count_cases (valexpstr, vallen, 0, &pars0);
case0n = count_cases (valexpstr, vallen, 1, &pars0);
vallen -= pars0;
case1 = count_cases (valexpstr, vallen, invert, &pars1);
vallen -= pars1;
case2 = count_cases (valexpstr, vallen, invert, &pars2);
*parsed = 1 + pars0 + pars1 + pars2;
// Recombine "eti?" as "et&ti&ei~&||"
// Note the test is not inverted, but the case's outcomes are
if ((case0p == -1) || (case0n == -1) || (case1 == -1) || (case2 == -1)) {
*opcp = ERR_TERNOP;
return -1;
}
*opcp = OPC_IF;
if (case1 == 0) {
*opcp |= zero_1st;
}
if (case2 == 0) {
*opcp |= zero_2nd;
}
retval = (case1 * case0p) + (case2 * case0n) + (case1 * case2);
return retval;
case ERR_BINOP:
*opcp = ERR_BINOP;
return -1;
case ERR_TERNOP:
*opcp = ERR_TERNOP;
return -1;
case '~':
case1 = count_cases (valexpstr, vallen, !invert, &pars1);
*parsed = 1 + pars1;
return case1;
case '0':
case '1':
*parsed = 1;
if (opc == (invert? '1': '0')) {
// strings like 0 and 1~ --> FALSE
// expands to no case at all
return 0;
} else {
// strings like 1 and 0~ --> TRUE
// expands to a single, already-fulfilled case
return 1;
}
default:
if (VALEXP_OPERAND (valexpstr [vallen])) {
*parsed = 1;
return 1;
} else {
// Syntax error
return -1;
}
}
}