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regexes, Slight refactor, but mostly fix comments,
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Also some white space changes.

This reinstates some refactoring that was done in the reverted commit
c29dfc6.  This was the part of the
commit that didn't lead to problems.  Instead, it makes the code
slightly simpler, with fewer instructions needed (disregarding any
compiler optimizations).

But mostly this commit rewords and amplifies some comments.  I
discovered I didn't fully understand things (as the algorithm was taken
from a book), and wrote the comments to be wrong or at least misleading.
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@khwilliamson
khwilliamson committed Jun 20, 2016
1 parent 18940b1 commit c04c2bf
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254 changes: 139 additions & 115 deletions regcomp.c
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Expand Up @@ -8951,10 +8951,12 @@ Perl__invlist_union_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,

/* running count, as explained in the algorithm source book; items are
* stopped accumulating and are output when the count changes to/from 0.
* The count is incremented when we start a range that's in the set, and
* decremented when we start a range that's not in the set. So its range
* is 0 to 2. Only when the count is zero is something not in the set.
*/
* The count is incremented when we start a range that's in an input's set,
* and decremented when we start a range that's not in a set. So this
* variable can be 0, 1, or 2. When it is 0 neither input is in their set,
* and hence nothing goes into the union; 1, just one of the inputs is in
* its set (and its current range gets added to the union); and 2 when both
* inputs are in their sets. */
UV count = 0;

PERL_ARGS_ASSERT__INVLIST_UNION_MAYBE_COMPLEMENT_2ND;
Expand Down Expand Up @@ -9017,7 +9019,7 @@ Perl__invlist_union_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,
invlist_replace_list_destroys_src(*output, u);
SvREFCNT_dec_NN(u);

return;
return;
}

if (a == NULL || ((len_a = _invlist_len(a)) == 0)) {
Expand Down Expand Up @@ -9088,32 +9090,32 @@ Perl__invlist_union_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,
u = _new_invlist(len_a + len_b);

/* Will contain U+0000 if either component does */
array_u = _invlist_array_init(u, (len_a > 0 && array_a[0] == 0)
|| (len_b > 0 && array_b[0] == 0));
array_u = _invlist_array_init(u, ( len_a > 0 && array_a[0] == 0)
|| (len_b > 0 && array_b[0] == 0));

/* Go through each list item by item, stopping when exhausted one of
/* Go through each input list item by item, stopping when exhausted one of
* them */
while (i_a < len_a && i_b < len_b) {
UV cp; /* The element to potentially add to the union's array */
bool cp_in_set; /* is it in the the input list's set or not */

/* We need to take one or the other of the two inputs for the union.
* Since we are merging two sorted lists, we take the smaller of the
* next items. In case of a tie, we take the one that is in its set
* first. If we took one not in the set first, it would decrement the
* count, possibly to 0 which would cause it to be output as ending the
* range, and the next time through we would take the same number, and
* output it again as beginning the next range. By doing it the
* opposite way, there is no possibility that the count will be
* momentarily decremented to 0, and thus the two adjoining ranges will
* be seamlessly merged. (In a tie and both are in the set or both not
* in the set, it doesn't matter which we take first.) */
if (array_a[i_a] < array_b[i_b]
|| (array_a[i_a] == array_b[i_b]
* next items. In case of a tie, we take first the one that is in its
* set. If we first took the one not in its set, it would decrement
* the count, possibly to 0 which would cause it to be output as ending
* the range, and the next time through we would take the same number,
* and output it again as beginning the next range. By doing it the
* opposite way, there is no possibility that the count will be
* momentarily decremented to 0, and thus the two adjoining ranges will
* be seamlessly merged. (In a tie and both are in the set or both not
* in the set, it doesn't matter which we take first.) */
if ( array_a[i_a] < array_b[i_b]
|| ( array_a[i_a] == array_b[i_b]
&& ELEMENT_RANGE_MATCHES_INVLIST(i_a)))
{
cp_in_set = ELEMENT_RANGE_MATCHES_INVLIST(i_a);
cp= array_a[i_a++];
cp = array_a[i_a++];
}
else {
cp_in_set = ELEMENT_RANGE_MATCHES_INVLIST(i_b);
Expand All @@ -9137,39 +9139,53 @@ Perl__invlist_union_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,
}
}

/* Here, we are finished going through at least one of the lists, which
* means there is something remaining in at most one. We check if the list
* that hasn't been exhausted is positioned such that we are in the middle
* of a range in its set or not. (i_a and i_b point to the element beyond
* the one we care about.) If in the set, we decrement 'count'; if 0, there
* is potentially more to output.
* There are four cases:
* 1) Both weren't in their sets, count is 0, and remains 0. What's left
* in the union is entirely from the non-exhausted set.
* 2) Both were in their sets, count is 2. Nothing further should
* be output, as everything that remains will be in the exhausted
* list's set, hence in the union; decrementing to 1 but not 0 insures
* that
* 3) the exhausted was in its set, non-exhausted isn't, count is 1.
* Nothing further should be output because the union includes
* everything from the exhausted set. Not decrementing ensures that.
* 4) the exhausted wasn't in its set, non-exhausted is, count is 1;
* decrementing to 0 insures that we look at the remainder of the
* non-exhausted set */

/* The loop above increments the index into exactly one of the input lists
* each iteration, and ends when either index gets to its list end. That
* means the other index is lower than its end, and so something is
* remaining in that one. We decrement 'count', as explained below, if
* that list is in its set. (i_a and i_b each currently index the element
* beyond the one we care about.) */
if ( (i_a != len_a && PREV_RANGE_MATCHES_INVLIST(i_a))
|| (i_b != len_b && PREV_RANGE_MATCHES_INVLIST(i_b)))
{
count--;
}

/* The final length is what we've output so far, plus what else is about to
* be output. (If 'count' is non-zero, then the input list we exhausted
* has everything remaining up to the machine's limit in its set, and hence
* in the union, so there will be no further output. */
len_u = i_u;
if (count == 0) {
/* At most one of the subexpressions will be non-zero */
len_u += (len_a - i_a) + (len_b - i_b);
/* Above we decremented 'count' if the list that had unexamined elements in
* it was in its set. This has made it so that 'count' being non-zero
* means there isn't anything left to output; and 'count' equal to 0 means
* that what is left to output is precisely that which is left in the
* non-exhausted input list.
*
* To see why, note first that the exhausted input obviously has nothing
* left to add to the union. If it was in its set at its end, that means
* the set extends from here to the platform's infinity, and hence so does
* the union and the non-exhausted set is irrelevant. The exhausted set
* also contributed 1 to 'count'. If 'count' was 2, it got decremented to
* 1, but if it was 1, the non-exhausted set wasn't in its set, and so
* 'count' remains at 1. This is consistent with the decremented 'count'
* != 0 meaning there's nothing left to add to the union.
*
* But if the exhausted input wasn't in its set, it contributed 0 to
* 'count', and the rest of the union will be whatever the other input is.
* If 'count' was 0, neither list was in its set, and 'count' remains 0;
* otherwise it gets decremented to 0. This is consistent with 'count'
* == 0 meaning the remainder of the union is whatever is left in the
* non-exhausted list. */
if (count != 0) {
len_u = i_u;
}
else {
IV copy_count = len_a - i_a;
if (copy_count > 0) { /* The non-exhausted input is 'a' */
Copy(array_a + i_a, array_u + i_u, copy_count, UV);
}
else { /* The non-exhausted input is b */
copy_count = len_b - i_b;
Copy(array_b + i_b, array_u + i_u, copy_count, UV);
}
len_u = i_u + copy_count;
}

/* Set the result to the final length, which can change the pointer to
Expand All @@ -9181,22 +9197,6 @@ Perl__invlist_union_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,
array_u = invlist_array(u);
}

/* When 'count' is 0, the list that was exhausted (if one was shorter than
* the other) ended with everything above it not in its set. That means
* that the remaining part of the union is precisely the same as the
* non-exhausted list, so can just copy it unchanged. (If both lists were
* exhausted at the same time, then the operations below will be both 0.)
*/
if (count == 0) {
IV copy_count; /* At most one will have a non-zero copy count */
if ((copy_count = len_a - i_a) > 0) {
Copy(array_a + i_a, array_u + i_u, copy_count, UV);
}
else if ((copy_count = len_b - i_b) > 0) {
Copy(array_b + i_b, array_u + i_u, copy_count, UV);
}
}

/* If the output is not to overwrite either of the inputs, just return the
* calculated union */
if (a != *output && b != *output) {
Expand All @@ -9209,7 +9209,7 @@ Perl__invlist_union_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,
* shown [perl #127392] that if the input is a mortal, we can get a
* huge build-up of these during regex compilation before they get
* freed. So for that case, replace just the input's interior with
* the output's, and then free the output */
* the union's, and then free the union */

assert(! invlist_is_iterating(*output));

Expand Down Expand Up @@ -9261,12 +9261,12 @@ Perl__invlist_intersection_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,
UV i_b = 0;
UV i_r = 0;

/* running count, as explained in the algorithm source book; items are
* stopped accumulating and are output when the count changes to/from 2.
* The count is incremented when we start a range that's in the set, and
* decremented when we start a range that's not in the set. So its range
* is 0 to 2. Only when the count is 2 is something in the intersection.
*/
/* running count of how many of the two inputs are postitioned at ranges
* that are in their sets. As explained in the algorithm source book,
* items are stopped accumulating and are output when the count changes
* to/from 2. The count is incremented when we start a range that's in an
* input's set, and decremented when we start a range that's not in a set.
* Only when it is 2 are we in the intersection. */
UV count = 0;

PERL_ARGS_ASSERT__INVLIST_INTERSECTION_MAYBE_COMPLEMENT_2ND;
Expand Down Expand Up @@ -9338,8 +9338,8 @@ Perl__invlist_intersection_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,
r= _new_invlist(len_a + len_b);

/* Will contain U+0000 iff both components do */
array_r = _invlist_array_init(r, len_a > 0 && array_a[0] == 0
&& len_b > 0 && array_b[0] == 0);
array_r = _invlist_array_init(r, len_a > 0 && array_a[0] == 0
&& len_b > 0 && array_b[0] == 0);

/* Go through each list item by item, stopping when exhausted one of
* them */
Expand All @@ -9350,21 +9350,21 @@ Perl__invlist_intersection_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,

/* We need to take one or the other of the two inputs for the
* intersection. Since we are merging two sorted lists, we take the
* smaller of the next items. In case of a tie, we take the one that
* is not in its set first (a difference from the union algorithm). If
* we took one in the set first, it would increment the count, possibly
* to 2 which would cause it to be output as starting a range in the
* intersection, and the next time through we would take that same
* number, and output it again as ending the set. By doing it the
* opposite of this, there is no possibility that the count will be
* momentarily incremented to 2. (In a tie and both are in the set or
* both not in the set, it doesn't matter which we take first.) */
if (array_a[i_a] < array_b[i_b]
|| (array_a[i_a] == array_b[i_b]
* smaller of the next items. In case of a tie, we take first the one
* that is not in its set (a difference from the union algorithm). If
* we first took the one in its set, it would increment the count,
* possibly to 2 which would cause it to be output as starting a range
* in the intersection, and the next time through we would take that
* same number, and output it again as ending the set. By doing the
* opposite of this, there is no possibility that the count will be
* momentarily incremented to 2. (In a tie and both are in the set or
* both not in the set, it doesn't matter which we take first.) */
if ( array_a[i_a] < array_b[i_b]
|| ( array_a[i_a] == array_b[i_b]
&& ! ELEMENT_RANGE_MATCHES_INVLIST(i_a)))
{
cp_in_set = ELEMENT_RANGE_MATCHES_INVLIST(i_a);
cp= array_a[i_a++];
cp = array_a[i_a++];
}
else {
cp_in_set = ELEMENT_RANGE_MATCHES_INVLIST(i_b);
Expand All @@ -9386,36 +9386,54 @@ Perl__invlist_intersection_maybe_complement_2nd(pTHX_ SV* const a, SV* const b,
}
count--;
}
}

/* Here, we are finished going through at least one of the lists, which
* means there is something remaining in at most one. We check if the list
* that has been exhausted is positioned such that we are in the middle
* of a range in its set or not. (i_a and i_b point to elements 1 beyond
* the ones we care about.) There are four cases:
* 1) Both weren't in their sets, count is 0, and remains 0. There's
* nothing left in the intersection.
* 2) Both were in their sets, count is 2 and perhaps is incremented to
* above 2. What should be output is exactly that which is in the
* non-exhausted set, as everything it has is also in the intersection
* set, and everything it doesn't have can't be in the intersection
* 3) The exhausted was in its set, non-exhausted isn't, count is 1, and
* gets incremented to 2. Like the previous case, the intersection is
* everything that remains in the non-exhausted set.
* 4) the exhausted wasn't in its set, non-exhausted is, count is 1, and
* remains 1. And the intersection has nothing more. */
}
/* The loop above increments the index into exactly one of the input lists
* each iteration, and ends when either index gets to its list end. That
* means the other index is lower than its end, and so something is
* remaining in that one. We increment 'count', as explained below, if the
* exhausted list was in its set. (i_a and i_b each currently index the element
* beyond the one we care about.) */
if ( (i_a == len_a && PREV_RANGE_MATCHES_INVLIST(i_a))
|| (i_b == len_b && PREV_RANGE_MATCHES_INVLIST(i_b)))
|| (i_b == len_b && PREV_RANGE_MATCHES_INVLIST(i_b)))
{
count++;
}

/* The final length is what we've output so far plus what else is in the
* intersection. At most one of the subexpressions below will be non-zero
* */
len_r = i_r;
if (count >= 2) {
len_r += (len_a - i_a) + (len_b - i_b);
/* Above we incremented 'count' if the exhausted list was in its set. This
* has made it so that 'count' being below 2 means there is nothing left to
* output; otheriwse what's left to add to the intersection is precisely
* that which is left in the non-exhausted input list.
*
* To see why, note first that the exhausted input obviously has nothing
* left to affect the intersection. If it was in its set at its end, that
* means the set extends from here to the platform's infinity, and hence
* anything in the non-exhausted's list will be in the intersection, and
* anything not in it won't be. Hence, the rest of the intersection is
* precisely what's in the non-exhausted list The exhausted set also
* contributed 1 to 'count', meaning 'count' was at least 1. Incrementing
* it means 'count' is now at least 2. This is consistent with the
* incremented 'count' being >= 2 means to add the non-exhausted list to
* the intersection.
*
* But if the exhausted input wasn't in its set, it contributed 0 to
* 'count', and the intersection can't include anything further; the
* non-exhausted set is irrelevant. 'count' was at most 1, and doesn't get
* incremented. This is consistent with 'count' being < 2 meaning nothing
* further to add to the intersection. */
if (count < 2) { /* Nothing left to put in the intersection. */
len_r = i_r;
}
else { /* copy the non-exhausted list, unchanged. */
IV copy_count = len_a - i_a;
if (copy_count > 0) { /* a is the one with stuff left */
Copy(array_a + i_a, array_r + i_r, copy_count, UV);
}
else { /* b is the one with stuff left */
copy_count = len_b - i_b;
Copy(array_b + i_b, array_r + i_r, copy_count, UV);
}
len_r = i_r + copy_count;
}

/* Set the result to the final length, which can change the pointer to
Expand Down Expand Up @@ -11351,7 +11369,7 @@ S_regbranch(pTHX_ RExC_state_t *pRExC_state, I32 *flagp, I32 first, U32 depth)
FAIL2("panic: regpiece returned NULL, flags=%#"UVxf"", (UV) flags);
}
else if (ret == NULL)
ret = latest;
ret = latest;
*flagp |= flags&(HASWIDTH|POSTPONED);
if (chain == NULL) /* First piece. */
*flagp |= flags&SPSTART;
Expand Down Expand Up @@ -12207,13 +12225,15 @@ S_backref_value(char *p)
/*
- regatom - the lowest level

Try to identify anything special at the start of the pattern. If there
is, then handle it as required. This may involve generating a single regop,
such as for an assertion; or it may involve recursing, such as to
handle a () structure.
Try to identify anything special at the start of the current parse position.
If there is, then handle it as required. This may involve generating a
single regop, such as for an assertion; or it may involve recursing, such as
to handle a () structure.

If the string doesn't start with something special then we gobble up
as much literal text as we can.
as much literal text as we can. If we encounter a quantifier, we have to
back off the final literal character, as that quantifier applies to just it
and not to the whole string of literals.

Once we have been able to handle whatever type of thing started the
sequence, we return.
Expand Down Expand Up @@ -12928,6 +12948,9 @@ S_regatom(pTHX_ RExC_state_t *pRExC_state, I32 *flagp, U32 depth)
|| UTF8_IS_INVARIANT(UCHARAT(RExC_parse))
|| UTF8_IS_START(UCHARAT(RExC_parse)));

/* Here, we have a literal character. Find the maximal string of
* them in the input that we can fit into a single EXACTish node.
* We quit at the first non-literal or when the node gets full */
for (p = RExC_parse;
len < upper_parse && p < RExC_end;
len++)
Expand Down Expand Up @@ -13223,6 +13246,7 @@ S_regatom(pTHX_ RExC_state_t *pRExC_state, I32 *flagp, U32 depth)
* the node, close the node with just them, and set up to do
* this character again next time through, when it will be the
* only thing in its new node */

if ((next_is_quantifier = ( LIKELY(p < RExC_end)
&& UNLIKELY(ISMULT2(p))))
&& LIKELY(len))
Expand Down Expand Up @@ -18107,7 +18131,7 @@ S_nextchar(pTHX_ RExC_state_t *pRExC_state)
RExC_parse += (UTF) ? UTF8SKIP(RExC_parse) : 1;

skip_to_be_ignored_text(pRExC_state, &RExC_parse,
FALSE /* Don't assume /x */ );
FALSE /* Don't force /x */ );
}
}

Expand Down

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