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Simplify lines consistently on opposite sides of tile boundaries

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This commit is contained in:
Eric Fischer
2016-01-07 11:35:11 -08:00
parent e1e028b865
commit 1f8b6faec8
4 changed files with 48 additions and 4 deletions
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tile.cc
+3 -3
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@@ -493,7 +493,7 @@ void *partial_feature_worker(void *v) {
for (unsigned i = a->task; i < (*partials).size(); i += a->tasks) {
drawvec geom = (*partials)[i].geom;
(*partials)[i].geom.clear(); // avoid keeping two copies in memory
(*partials)[i].geom.clear(); // avoid keeping two copies in memory
signed char t = (*partials)[i].t;
int z = (*partials)[i].z;
int line_detail = (*partials)[i].line_detail;
@@ -501,7 +501,7 @@ void *partial_feature_worker(void *v) {
char *additional = (*partials)[i].additional;
if ((t == VT_LINE || t == VT_POLYGON) && !prevent['s' & 0xFF]) {
if (1 /* !reduced */) { // XXX why did this not simplify if reduced?
if (1 /* !reduced */) { // XXX why did this not simplify if reduced?
if (t == VT_LINE) {
geom = remove_noop(geom, t, 32 - z - line_detail);
}
@@ -845,7 +845,7 @@ long long write_tile(char **geoms, char *metabase, char *stringpool, int z, unsi
// This is serial because decode_meta() unifies duplicates
for (unsigned i = 0; i < partials.size(); i++) {
drawvec geom = partials[i].geom;
partials[i].geom.clear(); // avoid keeping two copies in memory
partials[i].geom.clear(); // avoid keeping two copies in memory
long long layer = partials[i].layer;
char *meta = partials[i].meta;
signed char t = partials[i].t;