SplitApplication:
PROC [design:
CD.Design, aPtr:
CD.ApplicationPtr, from, to:
CD.DesignPosition] =
BEGIN
cutPos, oldRectPos, oldSize: CD.DesignPosition;
oldRect: CD.DesignRect; -- the bounding box
L1, L2, W1, W2: CD.DesignNumber;
wireLeft, wireRight: CD.ObPtr;
vertical, intersect: BOOL;
NearlyRoundToLambda:
PROC[x:
CD.DesignNumber]
RETURNS[r:
CD.DesignNumber] =
BEGIN
r ← (x/CD.lambda)*CD.lambda;
IF x=0 THEN r ← CD.lambda;
END;
FindCut:
PROC[from, to:
CD.DesignPosition, oldRect:
CD.DesignRect, vertical:
BOOL]
RETURNS[cutPos:
CD.DesignPosition, intersect:
BOOL ←
TRUE] =
BEGIN
For now OK, but check if there is any risk to make anything smaller than one lambda
inter, ctr: CD.DesignPosition; -- the intersection of the two lines, if any, and the center of the wire
dx, dy: CD.DesignNumber;
cursorBox: CD.DesignRect;
cursorBox ← CDInline.ToRect[from, to];
dx ← to.x-from.x; dy ← to.y-from.y;
ctr ← CDInline.Center[oldRect];
Test if vector parallel to wire
IF vertical AND (dx=0) OR (NOT vertical) AND (dy=0) THEN RETURN[[0,0], FALSE];
Now we can divide and find the intersection of lines (not yet segments)
inter ←
IF vertical
THEN [ctr.x, from.y+(dy*(ctr.x-from.x)/dx)]
ELSE [dx*(ctr.y-from.y)/dy+from.x, ctr.y];
Monier-Sindhu theorem: if inter is in cursorBox and in oldRect, then it is the intersection of segments
IF NOT (CDInline.InsidePos[inter, cursorBox] AND CDInline.InsidePos[inter, oldRect]) THEN RETURN[[0,0], FALSE];
cutPos ← IF vertical THEN [oldRect.x1, NearlyRoundToLambda[inter.y]] ELSE [NearlyRoundToLambda[inter.x], oldRect.y1];
END;
oldRect ← CDApplications.ARectI[aPtr];
oldRectPos ← CDInline.BaseOfRect[oldRect];
oldSize ← CDInline.SizeOfRect[oldRect];
vertical ← oldSize.y > oldSize.x;
[cutPos, intersect] ← FindCut[from, to, oldRect, vertical];
IF NOT intersect THEN RETURN;
IF vertical
THEN {
L1 ← cutPos.y - oldRect.y1;
L2 ← oldSize.y - L1;
W1 ← W2 ← oldSize.x;
}
ELSE {
L1 ← L2 ← oldSize.y;
W1 ← cutPos.x - oldRect.x1;
W2 ← oldSize.x - W1;
};
wireLeft ← CDRects.CreateRect[CDOrient.OrientedSize[[W1, L1], aPtr.orientation], aPtr.ob.level];
wireRight ← CDRects.CreateRect[CDOrient.OrientedSize[[W2, L2], aPtr.orientation], aPtr.ob.level];
CDOps.IncludeApplication[design, CDApplications.NewApplicationI[wireLeft, oldRectPos, aPtr.orientation, TRUE, CDProperties.CopyProps[aPtr.properties]], FALSE];
CDOps.IncludeApplication[design, CDApplications.NewApplicationI[wireRight, cutPos, aPtr.orientation, TRUE, CDProperties.CopyProps[aPtr.properties]], FALSE];
CDOps.RemoveApplication[design, aPtr];
END;
DoAllSplits:
PROC [design:
CD.Design, from, to:
CD.DesignPosition, pointed:
BOOL←
FALSE] =
BEGIN
The mouse track defines the cut vector. Let's define the "spine" of a wire as the centerline, roughly in the middle (lambda) and following the visible length, i.e. the longer edge, not the CD length. Then we cut the wire at the position where the spine meets the cut vector. The cut always reduces the visible length of the wire. (Pff, not easy to describe without drawing!)
A vertical wire freshly drawn has even (0) orientation.
FOR list:
CD.ApplicationList ← CDOps.AppList[design], list.rest
WHILE list#
NIL
DO
IF list.first.selected AND list.first.ob.p.wireTyped THEN SplitApplication[design, list.first, from, to];
ENDLOOP;
END;