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Use < instead of <= when comparing against TOLERANCE
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2 changed files with 6 additions and 6 deletions
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@ -1049,7 +1049,7 @@ new function() { // Scope for methods that require numerical integration
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// Get length of total range
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rangeLength = Numerical.integrate(ds, a, b,
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getIterations(a, b));
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if (abs(offset - rangeLength) <= tolerance) {
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if (abs(offset - rangeLength) < tolerance) {
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// Matched the end:
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return forward ? b : a;
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} else if (abs(offset) > rangeLength) {
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@ -1115,7 +1115,7 @@ new function() { // Scope for methods that require numerical integration
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dp2 = getSignedDistance(q0x, q0y, q3x, q3y, v1[4], v1[5]),
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dp3 = getSignedDistance(q0x, q0y, q3x, q3y, v1[6], v1[7]),
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tMinNew, tMaxNew, tDiff;
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if (q0x === q3x && uMax - uMin <= tolerance && recursion > 3) {
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if (q0x === q3x && uMax - uMin < tolerance && recursion > 3) {
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// The fatline of Q has converged to a point, the clipping is not
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// reliable. Return the value we have even though we will miss the
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// precision.
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@ -138,15 +138,15 @@ PathItem.inject(new function() {
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if (length <= curveLength) {
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// If the selected location on the curve falls onto its
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// beginning or end, use the curve's center instead.
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if (length <= tolerance
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|| curveLength - length <= tolerance)
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if (length < tolerance
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|| curveLength - length < tolerance)
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length = curveLength / 2;
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var curve = node.segment.getCurve(),
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pt = curve.getPointAt(length),
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// Determine if the curve is a horizontal linear
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// curve by checking the slope of it's tangent.
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hor = curve.isLinear() && Math.abs(curve
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.getTangentAt(0.5, true).y) <= tolerance,
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.getTangentAt(0.5, true).y) < tolerance,
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path = curve._path;
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if (path._parent instanceof CompoundPath)
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path = path._parent;
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@ -320,7 +320,7 @@ PathItem.inject(new function() {
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// Detect and exclude intercepts at 'end' of loops:
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&& (i === l - 1 || curve.next !== curves[i + 1])
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&& abs(Curve.evaluate(curve.next.values, 0, 0).x -x)
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<= tolerance
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< tolerance
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// Detect 2nd case of a consecutive intercept, but make
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// sure we're still on the same loop
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|| i > 0 && curve.previous === curves[i - 1]
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