zdoom
/
qzdoom
mirror of https://github.com/ZDoom/qzdoom.git
0
0
Fork 0
Code Issues Packages Projects Releases Wiki Activity

Merge branch 'slope64_comment' of https://github.com/edward-san/zdoom

This commit is contained in:
Christoph Oelckers 2014-11-25 17:39:08 +01:00
parent 29ecbac963 cc4e66f976
commit a67ac5d940
1 changed files with 44 additions and 12 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

56
src/p_map.cpp
View File

@ -737,18 +737,50 @@ bool PIT_CheckLine(line_t *ld, const FBoundingBox &box, FCheckPosition &tm)
{ // Find the point on the line closest to the actor's center, and use
// that to calculate openings
// [EP] Use 64 bit integers in order to keep the exact result of the
// multiplication, because in the worst case, which is by the map limit
// (32767 units, which is 2147418112 in fixed_t notation), the result
// would occupy 62 bits (if I consider also the addition with another
// and possible 62 bit value, it's 63 bits).
// This privilege could not be available if the starting data would be
// 64 bit long.
// With this, the division is exact as the 32 bit float counterpart,
// though I don't know why I had to discard the first 24 bits from the
// divisor.
SQWORD r_num = ((SQWORD(tm.x - ld->v1->x)*ld->dx) + (SQWORD(tm.y - ld->v1->y)*ld->dy));
SQWORD r_den = (SQWORD(ld->dx)*ld->dx + SQWORD(ld->dy)*ld->dy) / (1 << 24);
fixed_t r = (fixed_t)(r_num / r_den);
// multiplication, because in the case the vertexes have both the
// distance coordinates equal to the map limit (32767 units, which is
// 2147418112 in fixed_t notation), the product result would occupy
// 62 bits and the sum of two products would occupy 63 bits
// in the worst case. If instead the vertexes are very close (1 in
// fixed_t notation, which is 1.52587890625e-05 in float notation), the
// product and the sum can be 1 in the worst case, which is very tiny.
SQWORD r_num = ((SQWORD(tm.x - ld->v1->x)*ld->dx) +
(SQWORD(tm.y - ld->v1->y)*ld->dy));
// The denominator is always positive. Use this to avoid useless
// calculations.
SQWORD r_den = (SQWORD(ld->dx)*ld->dx + SQWORD(ld->dy)*ld->dy);
fixed_t r = 0;
if (r_num <= 0) {
// [EP] The numerator is less or equal to zero, hence the closest
// point on the line is the first vertex. Truncate the result to 0.
r = 0;
} else if (r_num >= r_den) {
// [EP] The division is greater or equal to 1, hence the closest
// point on the line is the second vertex. Truncate the result to
// 1 << 24.
r = (1 << 24);
} else {
// [EP] Deal with the limited bits. The original formula is:
// r = (r_num << 24) / r_den,
// but r_num might be big enough to make the shift overflow.
// Since the numerator can't be saved in a 128bit integer,
// the denominator must be right shifted. If the denominator is
// less than (1 << 24), there would be a division by zero.
// Thanks to the fact that in this code path the denominator is less
// than the numerator, it's possible to avoid this bad situation by
// just checking the last 24 bits of the numerator.
if ((r_num >> (63-24)) != 0) {
// [EP] In fact, if the numerator is greater than
// (1 << (63-24)), the denominator must be greater than
// (1 << (63-24)), hence the denominator won't be zero after
// the right shift by 24 places.
r = (r_num)/(r_den >> 24);
} else {
// [EP] Having the last 24 bits all zero allows right shifting
// the numerator by 24 bits.
r = (r_num << 24)/r_den;
}
}
/* Printf ("%d:%d: %d (%d %d %d %d) (%d %d %d %d)\n", level.time, ld-lines, r,
ld->frontsector->floorplane.a,
ld->frontsector->floorplane.b,