本文整理汇总了C++中SkDLine::xyAtT方法的典型用法代码示例。如果您正苦于以下问题：C++ SkDLine::xyAtT方法的具体用法？C++ SkDLine::xyAtT怎么用？C++ SkDLine::xyAtT使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类SkDLine的用法示例。
在下文中一共展示了SkDLine::xyAtT方法的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: computePoints
int SkIntersections::computePoints(const SkDLine& line, int used) {
fPt[0] = line.xyAtT(fT[0][0]);
if ((fUsed = used) == 2) {
fPt[1] = line.xyAtT(fT[0][1]);
}
return fUsed;
}示例2: testLineIntersect
static void testLineIntersect(skiatest::Reporter* reporter, const SkDQuad& quad,
const SkDLine& line, const double x, const double y) {
char pathStr[1024];
sk_bzero(pathStr, sizeof(pathStr));
char* str = pathStr;
str += sprintf(str, " path.moveTo(%1.9g, %1.9g);\n", quad[0].fX, quad[0].fY);
str += sprintf(str, " path.quadTo(%1.9g, %1.9g, %1.9g, %1.9g);\n", quad[1].fX,
quad[1].fY, quad[2].fX, quad[2].fY);
str += sprintf(str, " path.moveTo(%1.9g, %1.9g);\n", line[0].fX, line[0].fY);
str += sprintf(str, " path.lineTo(%1.9g, %1.9g);\n", line[1].fX, line[1].fY);
SkIntersections intersections;
bool flipped = false;
int result = doIntersect(intersections, quad, line, flipped);
bool found = false;
for (int index = 0; index < result; ++index) {
double quadT = intersections[0][index];
SkDPoint quadXY = quad.xyAtT(quadT);
double lineT = intersections[1][index];
SkDPoint lineXY = line.xyAtT(lineT);
if (quadXY.approximatelyEqual(lineXY)) {
found = true;
}
}
REPORTER_ASSERT(reporter, found);
}示例3: intersect
// note that this only works if both lines are neither horizontal nor vertical
int SkIntersections::intersect(const SkDLine& a, const SkDLine& b) {
// see if end points intersect the opposite line
double t;
for (int iA = 0; iA < 2; ++iA) {
if (!checkEndPoint(a[iA].fX, a[iA].fY, b, &t, -1)) {
continue;
}
insert(iA, t, a[iA]);
}
for (int iB = 0; iB < 2; ++iB) {
if (!checkEndPoint(b[iB].fX, b[iB].fY, a, &t, -1)) {
continue;
}
insert(t, iB, b[iB]);
}
if (used() > 0) {
SkASSERT(fUsed <= 2);
return used(); // coincident lines are returned here
}
/* Determine the intersection point of two line segments
Return FALSE if the lines don't intersect
from: http://paulbourke.net/geometry/lineline2d/ */
double axLen = a[1].fX - a[0].fX;
double ayLen = a[1].fY - a[0].fY;
double bxLen = b[1].fX - b[0].fX;
double byLen = b[1].fY - b[0].fY;
/* Slopes match when denom goes to zero:
axLen / ayLen == bxLen / byLen
(ayLen * byLen) * axLen / ayLen == (ayLen * byLen) * bxLen / byLen
byLen * axLen == ayLen * bxLen
byLen * axLen - ayLen * bxLen == 0 ( == denom )
*/
double denom = byLen * axLen - ayLen * bxLen;
double ab0y = a[0].fY - b[0].fY;
double ab0x = a[0].fX - b[0].fX;
double numerA = ab0y * bxLen - byLen * ab0x;
double numerB = ab0y * axLen - ayLen * ab0x;
bool mayNotOverlap = (numerA < 0 && denom > numerA) || (numerA > 0 && denom < numerA)
|| (numerB < 0 && denom > numerB) || (numerB > 0 && denom < numerB);
numerA /= denom;
numerB /= denom;
if ((!approximately_zero(denom) || (!approximately_zero_inverse(numerA)
&& !approximately_zero_inverse(numerB))) && !sk_double_isnan(numerA)
&& !sk_double_isnan(numerB)) {
if (mayNotOverlap) {
return 0;
}
fT[0][0] = numerA;
fT[1][0] = numerB;
fPt[0] = a.xyAtT(numerA);
return computePoints(a, 1);
}
return 0;
}示例4: intersect
int SkIntersections::intersect(const SkDLine& a, const SkDLine& b) {
double axLen = a[1].fX - a[0].fX;
double ayLen = a[1].fY - a[0].fY;
double bxLen = b[1].fX - b[0].fX;
double byLen = b[1].fY - b[0].fY;
/* Slopes match when denom goes to zero:
axLen / ayLen == bxLen / byLen
(ayLen * byLen) * axLen / ayLen == (ayLen * byLen) * bxLen / byLen
byLen * axLen == ayLen * bxLen
byLen * axLen - ayLen * bxLen == 0 ( == denom )
*/
double denom = byLen * axLen - ayLen * bxLen;
double ab0y = a[0].fY - b[0].fY;
double ab0x = a[0].fX - b[0].fX;
double numerA = ab0y * bxLen - byLen * ab0x;
double numerB = ab0y * axLen - ayLen * ab0x;
bool mayNotOverlap = (numerA < 0 && denom > numerA) || (numerA > 0 && denom < numerA)
|| (numerB < 0 && denom > numerB) || (numerB > 0 && denom < numerB);
numerA /= denom;
numerB /= denom;
if ((!approximately_zero(denom) || (!approximately_zero_inverse(numerA)
&& !approximately_zero_inverse(numerB))) && !sk_double_isnan(numerA)
&& !sk_double_isnan(numerB)) {
if (mayNotOverlap) {
return fUsed = 0;
}
fT[0][0] = numerA;
fT[1][0] = numerB;
fPt[0] = a.xyAtT(numerA);
return computePoints(a, 1);
}
/* See if the axis intercepts match:
ay - ax * ayLen / axLen == by - bx * ayLen / axLen
axLen * (ay - ax * ayLen / axLen) == axLen * (by - bx * ayLen / axLen)
axLen * ay - ax * ayLen == axLen * by - bx * ayLen
*/
if (!AlmostEqualUlps(axLen * a[0].fY - ayLen * a[0].fX,
axLen * b[0].fY - ayLen * b[0].fX)) {
return fUsed = 0;
}
const double* aPtr;
const double* bPtr;
if (fabs(axLen) > fabs(ayLen) || fabs(bxLen) > fabs(byLen)) {
aPtr = &a[0].fX;
bPtr = &b[0].fX;
} else {
aPtr = &a[0].fY;
bPtr = &b[0].fY;
}
double a0 = aPtr[0];
double a1 = aPtr[2];
double b0 = bPtr[0];
double b1 = bPtr[2];
// OPTIMIZATION: restructure to reject before the divide
// e.g., if ((a0 - b0) * (a0 - a1) < 0 || abs(a0 - b0) > abs(a0 - a1))
// (except efficient)
double aDenom = a0 - a1;
if (approximately_zero(aDenom)) {
if (!between(b0, a0, b1)) {
return fUsed = 0;
}
fT[0][0] = fT[0][1] = 0;
} else {
double at0 = (a0 - b0) / aDenom;
double at1 = (a0 - b1) / aDenom;
if ((at0 < 0 && at1 < 0) || (at0 > 1 && at1 > 1)) {
return fUsed = 0;
}
fT[0][0] = SkTMax(SkTMin(at0, 1.0), 0.0);
fT[0][1] = SkTMax(SkTMin(at1, 1.0), 0.0);
}
double bDenom = b0 - b1;
if (approximately_zero(bDenom)) {
fT[1][0] = fT[1][1] = 0;
} else {
int bIn = aDenom * bDenom < 0;
fT[1][bIn] = SkTMax(SkTMin((b0 - a0) / bDenom, 1.0), 0.0);
fT[1][!bIn] = SkTMax(SkTMin((b0 - a1) / bDenom, 1.0), 0.0);
}
bool second = fabs(fT[0][0] - fT[0][1]) > FLT_EPSILON;
SkASSERT((fabs(fT[1][0] - fT[1][1]) <= FLT_EPSILON) ^ second);
return computePoints(a, 1 + second);
}