use double to compute root to avoid overflow
Bug: 850350
Change-Id: Iac04fc62e69f51b68c5fc7f55ac1be930133cc74
Reviewed-on: https://skia-review.googlesource.com/136597
Reviewed-by: Mike Reed <reed@google.com>
Commit-Queue: Mike Reed <reed@google.com>
diff --git a/src/core/SkGeometry.cpp b/src/core/SkGeometry.cpp
index a534509..ace8a7d 100644
--- a/src/core/SkGeometry.cpp
+++ b/src/core/SkGeometry.cpp
@@ -60,6 +60,15 @@
return 1;
}
+// Just returns its argument, but makes it easy to set a break-point to know when
+// SkFindUnitQuadRoots is going to return 0 (an error).
+static int return_check_zero(int value) {
+ if (value == 0) {
+ return 0;
+ }
+ return value;
+}
+
/** From Numerical Recipes in C.
Q = -1/2 (B + sign(B) sqrt[B*B - 4*A*C])
@@ -70,22 +79,21 @@
SkASSERT(roots);
if (A == 0) {
- return valid_unit_divide(-C, B, roots);
+ return return_check_zero(valid_unit_divide(-C, B, roots));
}
SkScalar* r = roots;
- SkScalar R = B*B - 4*A*C;
- if (R < 0 || !SkScalarIsFinite(R)) { // complex roots
- // if R is infinite, it's possible that it may still produce
- // useful results if the operation was repeated in doubles
- // the flipside is determining if the more precise answer
- // isn't useful because surrounding machinery (e.g., subtracting
- // the axis offset from C) already discards the extra precision
- // more investigation and unit tests required...
- return 0;
+ // use doubles so we don't overflow temporarily trying to compute R
+ double dr = (double)B * B - 4 * (double)A * C;
+ if (dr < 0) {
+ return return_check_zero(0);
}
- R = SkScalarSqrt(R);
+ dr = sqrt(dr);
+ SkScalar R = SkDoubleToScalar(dr);
+ if (!SkScalarIsFinite(R)) {
+ return return_check_zero(0);
+ }
SkScalar Q = (B < 0) ? -(B-R)/2 : -(B+R)/2;
r += valid_unit_divide(Q, A, r);
@@ -98,7 +106,7 @@
r -= 1; // skip the double root
}
}
- return (int)(r - roots);
+ return return_check_zero((int)(r - roots));
}
///////////////////////////////////////////////////////////////////////////////
