ralloc heap) shift the data out from underneath the regexp
routines.
- Here's another reason to avoid allocation: Emacs insists on
- processing input from X in a signal handler; processing X input may
+ Here's another reason to avoid allocation: Emacs
+ processes input from X in a signal handler; processing X input may
call malloc; if input arrives while a matching routine is calling
malloc, then we're scrod. But Emacs can't just block input while
calling matching routines; then we don't notice interrupts when
/* Normally, this is fine. */
#define MATCH_MAY_ALLOCATE
-/* But under some circumstances, it's not. */
-#if defined (emacs) || (defined (REL_ALLOC) && defined (C_ALLOCA))
+/* The match routines may not allocate if (1) they would do it with malloc
+ and (2) it's not safe for htem to use malloc. */
+#if (defined (C_ALLOCA) || defined (REGEX_MALLOC)) && (defined (emacs) || defined (REL_ALLOC))
#undef MATCH_MAY_ALLOCATE
#endif
/* Return, freeing storage we allocated. */
#define FREE_STACK_RETURN(value) \
-do \
-{ \
- free (compile_stack.stack); \
- return value; \
-} \
-while (1)
+ return (free (compile_stack.stack), value)
static reg_errcode_t
regex_compile (pattern, size, syntax, bufp)
We also want to fetch the endpoints without translating them; the
appropriate translation is done in the bit-setting loop below. */
- range_start = ((unsigned char *) p)[-2];
- range_end = ((unsigned char *) p)[0];
+ range_start = ((unsigned const char *) p)[-2];
+ range_end = ((unsigned const char *) p)[0];
/* Have to increment the pointer into the pattern string, so the
caller isn't still at the ending character. */
#endif
if ((re_opcode_t) p1[3] == exactn
- && ! (p2[1] * BYTEWIDTH > p1[4]
+ && ! ((int) p2[1] * BYTEWIDTH > (int) p1[4]
&& (p2[1 + p1[4] / BYTEWIDTH]
& (1 << (p1[4] % BYTEWIDTH)))))
{
int idx;
/* We win if the charset_not inside the loop
lists every character listed in the charset after. */
- for (idx = 0; idx < p2[1]; idx++)
+ for (idx = 0; idx < (int) p2[1]; idx++)
if (! (p2[2 + idx] == 0
- || (idx < p1[4]
+ || (idx < (int) p1[4]
&& ((p2[2 + idx] & ~ p1[5 + idx]) == 0))))
break;
int idx;
/* We win if the charset inside the loop
has no overlap with the one after the loop. */
- for (idx = 0; idx < p2[1] && idx < p1[4]; idx++)
+ for (idx = 0;
+ idx < (int) p2[1] && idx < (int) p1[4];
+ idx++)
if ((p2[2 + idx] & p1[5 + idx]) != 0)
break;