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1111 lines
37 KiB
1111 lines
37 KiB
/*************************************************************************\
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* Module Name: EngLine.cxx
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*
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* EngLine for bitmap simulations
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*
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* Created: 5-Apr-91
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* Author: Paul Butzi
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*
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* Copyright (c) 1990-1999 Microsoft Corporation
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\**************************************************************************/
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#include "precomp.hxx"
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#include "engline.hxx"
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/******************************Public*Routine******************************\
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* BOOL bLines(pbmi, pptfxFirst, pptfxBuf, prun, cptfx, pls, prclClip,
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* apfn[], flStart, pchBits, lDelta)
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*
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* Computes the DDA for the line and gets ready to draw it. Puts the
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* pixel data into an array of strips, and calls a strip routine to
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* do the actual drawing.
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*
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* Doing Lines Right
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* -----------------
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*
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* In NT, all lines are given to the device driver in fractional
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* coordinates, in a 28.4 fixed point format. The lower 4 bits are
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* fractional for sub-pixel positioning.
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*
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* Note that you CANNOT! just round the coordinates to integers
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* and pass the results to your favorite integer Bresenham routine!!
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* (Unless, of course, you have such a high resolution device that
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* nobody will notice -- not likely for a display device.) The
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* fractions give a more accurate rendering of the line -- this is
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* important for things like our Bezier curves, which would have 'kinks'
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* if the points in its polyline approximation were rounded to integers.
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*
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* Unfortunately, for fractional lines there is more setup work to do
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* a DDA than for integer lines. However, the main loop is exactly
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* the same (and can be done entirely with 32 bit math).
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*
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* Our Implementation
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* ------------------
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*
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* We employ a run length slice algorithm: our DDA calculates the
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* number of pixels that are in each row (or 'strip') of pixels.
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*
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* We've separated the running of the DDA and the drawing of pixels:
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* we run the DDA for several iterations and store the results in
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* a 'strip' buffer (which are the lengths of consecutive pixel rows of
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* the line), then we crank up a 'strip drawer' that will draw all the
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* strips in the buffer.
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*
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* We also employ a 'half-flip' to reduce the number of strip
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* iterations we need to do in the DDA and strip drawing loops: when a
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* (normalized) line's slope is more than 1/2, we do a final flip
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* about the line y = (1/2)x. So now, instead of each strip being
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* consecutive horizontal or vertical pixel rows, each strip is composed
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* of those pixels aligned in 45 degree rows. So a line like (0, 0) to
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* (128, 128) would generate only one strip.
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*
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* We also always draw only left-to-right.
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*
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* Style lines may have arbitrary style patterns. We specially
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* optimize the default patterns (and call them 'masked' styles).
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*
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* The DDA Derivation
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* ------------------
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*
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* Here is how I like to think of the DDA calculation.
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*
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* We employ Knuth's "diamond rule": rendering a one-pixel-wide line
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* can be thought of as dragging a one-pixel-wide by one-pixel-high
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* diamond along the true line. Pixel centers lie on the integer
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* coordinates, and so we light any pixel whose center gets covered
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* by the "drag" region (John D. Hobby, Journal of the Association
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* for Computing Machinery, Vol. 36, No. 2, April 1989, pp. 209-229).
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*
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* We must define which pixel gets lit when the true line falls
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* exactly half-way between two pixels. In this case, we follow
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* the rule: when two pels are equidistant, the upper or left pel
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* is illuminated, unless the slope is exactly one, in which case
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* the upper or right pel is illuminated. (So we make the edges
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* of the diamond exclusive, except for the top and left vertices,
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* which are inclusive, unless we have slope one.)
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*
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* This metric decides what pixels should be on any line BEFORE it is
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* flipped around for our calculation. Having a consistent metric
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* this way will let our lines blend nicely with our curves. The
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* metric also dictates that we will never have one pixel turned on
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* directly above another that's turned on. We will also never have
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* a gap; i.e., there will be exactly one pixel turned on for each
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* column between the start and end points. All that remains to be
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* done is to decide how many pixels should be turned on for each row.
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*
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* So lines we draw will consist of varying numbers of pixels on
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* successive rows, for example:
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*
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* ******
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* *****
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* ******
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* *****
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*
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* We'll call each set of pixels on a row a "strip".
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*
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* (Please remember that our coordinate space has the origin as the
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* upper left pixel on the screen; postive y is down and positive x
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* is right.)
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*
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* Device coordinates are specified as fixed point 28.4 numbers,
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* where the first 28 bits are the integer coordinate, and the last
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* 4 bits are the fraction. So coordinates may be thought of as
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* having the form (x, y) = (M/F, N/F) where F is the constant scaling
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* factor F = 2^4 = 16, and M and N are 32 bit integers.
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*
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* Consider the line from (M0/F, N0/F) to (M1/F, N1/F) which runs
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* left-to-right and whose slope is in the first octant, and let
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* dM = M1 - M0 and dN = N1 - N0. Then dM >= 0, dN >= 0 and dM >= dN.
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*
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* Since the slope of the line is less than 1, the edges of the
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* drag region are created by the top and bottom vertices of the
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* diamond. At any given pixel row y of the line, we light those
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* pixels whose centers are between the left and right edges.
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*
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* Let mL(n) denote the line representing the left edge of the drag
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* region. On pixel row j, the column of the first pixel to be
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* lit is
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*
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* iL(j) = ceiling( mL(j * F) / F)
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*
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* Since the line's slope is less than one:
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*
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* iL(j) = ceiling( mL([j + 1/2] F) / F )
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*
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* Recall the formula for our line:
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*
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* n(m) = (dN / dM) (m - M0) + N0
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*
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* m(n) = (dM / dN) (n - N0) + M0
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*
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* Since the line's slope is less than one, the line representing
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* the left edge of the drag region is the original line offset
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* by 1/2 pixel in the y direction:
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*
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* mL(n) = (dM / dN) (n - F/2 - N0) + M0
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*
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* From this we can figure out the column of the first pixel that
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* will be lit on row j, being careful of rounding (if the left
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* edge lands exactly on an integer point, the pixel at that
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* point is not lit because of our rounding convention):
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*
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* iL(j) = floor( mL(j F) / F ) + 1
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*
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* = floor( ((dM / dN) (j F - F/2 - N0) + M0) / F ) + 1
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*
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* = floor( F dM j - F/2 dM - N0 dM + dN M0) / F dN ) + 1
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*
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* F dM j - [ dM (N0 + F/2) - dN M0 ]
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* = floor( ---------------------------------- ) + 1
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* F dN
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*
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* dM j - [ dM (N0 + F/2) - dN M0 ] / F
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* = floor( ------------------------------------ ) + 1 (1)
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* dN
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*
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* = floor( (dM j + alpha) / dN ) + 1
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*
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* where
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*
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* alpha = - [ dM (N0 + F/2) - dN M0 ] / F
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*
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* We use equation (1) to calculate the DDA: there are iL(j+1) - iL(j)
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* pixels in row j. Because we are always calculating iL(j) for
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* integer quantities of j, we note that the only fractional term
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* is constant, and so we can 'throw away' the fractional bits of
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* alpha:
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*
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* beta = floor( - [ dM (N0 + F/2) - dN M0 ] / F ) (2)
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*
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* so
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*
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* iL(j) = floor( (dM j + beta) / dN ) + 1 (3)
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*
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* for integers j.
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*
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* Note if iR(j) is the line's rightmost pixel on row j, that
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* iR(j) = iL(j + 1) - 1.
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*
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* Similarly, rewriting equation (1) as a function of column i,
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* we can determine, given column i, on which pixel row j is the line
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* lit:
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*
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* dN i + [ dM (N0 + F/2) - dN M0 ] / F
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* j(i) = ceiling( ------------------------------------ ) - 1
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* dM
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*
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* Floors are easier to compute, so we can rewrite this:
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*
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* dN i + [ dM (N0 + F/2) - dN M0 ] / F + dM - 1/F
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* j(i) = floor( ----------------------------------------------- ) - 1
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* dM
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*
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* dN i + [ dM (N0 + F/2) - dN M0 ] / F + dM - 1/F - dM
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* = floor( ---------------------------------------------------- )
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* dM
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*
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* dN i + [ dM (N0 + F/2) - dN M0 - 1 ] / F
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* = floor( ---------------------------------------- )
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* dM
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*
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* We can once again wave our hands and throw away the fractional bits
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* of the remainder term:
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*
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* j(i) = floor( (dN i + gamma) / dM ) (4)
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*
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* where
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*
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* gamma = floor( [ dM (N0 + F/2) - dN M0 - 1 ] / F ) (5)
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*
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* We now note that
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*
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* beta = -gamma - 1 = ~gamma (6)
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*
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* To draw the pixels of the line, we could evaluate (3) on every scan
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* line to determine where the strip starts. Of course, we don't want
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* to do that because that would involve a multiply and divide for every
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* scan. So we do everything incrementally.
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*
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* We would like to easily compute c , the number of pixels on scan j:
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* j
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*
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* c = iL(j + 1) - iL(j)
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* j
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*
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* = floor((dM (j + 1) + beta) / dN) - floor((dM j + beta) / dN) (7)
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*
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* This may be rewritten as
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*
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* c = floor(i + r / dN) - floor(i + r / dN) (8)
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* j j+1 j+1 j j
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*
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* where i , i are integers and r < dN, r < dN.
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* j j+1 j j+1
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*
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* Rewriting (7) again:
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*
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* c = floor(i + r / dN + dM / dN) - floor(i + r / dN)
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* j j j j j
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*
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*
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* = floor((r + dM) / dN) - floor(r / dN)
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* j j
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*
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* This may be rewritten as
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*
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* c = dI + floor((r + dR) / dN) - floor(r / dN)
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* j j j
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*
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* where dI + dR / dN = dM / dN, dI is an integer and dR < dN.
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*
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* r is the remainder (or "error") term in the DDA loop: r / dN
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* j j
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* is the exact fraction of a pixel at which the strip ends. To go
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* on to the next scan and compute c we need to know r .
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* j+1 j+1
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*
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* So in the main loop of the DDA:
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*
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* c = dI + floor((r + dR) / dN) and r = (r + dR) % dN
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* j j j+1 j
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*
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* and we know r < dN, r < dN, and dR < dN.
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* j j+1
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*
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* We have derived the DDA only for lines in the first octant; to
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* handle other octants we do the common trick of flipping the line
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* to the first octant by first making the line left-to-right by
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* exchanging the end-points, then flipping about the lines y = 0 and
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* y = x, as necessary. We must record the transformation so we can
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* undo them later.
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*
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* We must also be careful of how the flips affect our rounding. If
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* to get the line to the first octant we flipped about x = 0, we now
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* have to be careful to round a y value of 1/2 up instead of down as
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* we would for a line originally in the first octant (recall that
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* "In the case where two pels are equidistant, the upper or left
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* pel is illuminated...").
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*
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* To account for this rounding when running the DDA, we shift the line
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* (or not) in the y direction by the smallest amount possible. That
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* takes care of rounding for the DDA, but we still have to be careful
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* about the rounding when determining the first and last pixels to be
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* lit in the line.
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*
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* Determining The First And Last Pixels In The Line
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* -------------------------------------------------
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*
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* Fractional coordinates also make it harder to determine which pixels
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* will be the first and last ones in the line. We've already taken
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* the fractional coordinates into account in calculating the DDA, but
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* the DDA cannot tell us which are the end pixels because it is quite
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* happy to calculate pixels on the line from minus infinity to positive
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* infinity.
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*
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* The diamond rule determines the start and end pixels. (Recall that
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* the sides are exclusive except for the left and top vertices.)
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* This convention can be thought of in another way: there are diamonds
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* around the pixels, and wherever the true line crosses a diamond,
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* that pel is illuminated.
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*
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* Consider a line where we've done the flips to the first octant, and the
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* floor of the start coordinates is the origin:
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*
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* +-----------------------> +x
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* |
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* | 0 1
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* | 0123456789abcdef
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* |
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* | 0 00000000?1111111
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* | 1 00000000 1111111
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* | 2 0000000 111111
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* | 3 000000 11111
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* | 4 00000 ** 1111
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* | 5 0000 ****1
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* | 6 000 1***
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* | 7 00 1 ****
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* | 8 ? ***
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* | 9 22 3 ****
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* | a 222 33 ***
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* | b 2222 333 ****
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* | c 22222 3333 **
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* | d 222222 33333
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* | e 2222222 333333
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* | f 22222222 3333333
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* |
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* | 2 3
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* v
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* +y
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*
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* If the start of the line lands on the diamond around pixel 0 (shown by
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* the '0' region here), pixel 0 is the first pel in the line. The same
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* is true for the other pels.
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*
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* A little more work has to be done if the line starts in the
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* 'nether-land' between the diamonds (as illustrated by the '*' line):
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* the first pel lit is the first diamond crossed by the line (pixel 1 in
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* our example). This calculation is determined by the DDA or slope of
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* the line.
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*
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* If the line starts exactly half way between two adjacent pixels
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* (denoted here by the '?' spots), the first pixel is determined by our
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* round-down convention (and is dependent on the flips done to
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* normalize the line).
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*
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* Last Pel Exclusive
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* ------------------
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*
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* To eliminate repeatedly lit pels between continuous connected lines,
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* we employ a last-pel exclusive convention: if the line ends exactly on
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* the diamond around a pel, that pel is not lit. (This eliminates the
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* checks we had in the old code to see if we were re-lighting pels.)
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*
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* The Half Flip
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* -------------
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*
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* To make our run length algorithm more efficient, we employ a "half
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* flip". If after normalizing to the first octant, the slope is more
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* than 1/2, we subtract the y coordinate from the x coordinate. This
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* has the effect of reflecting the coordinates through the line of slope
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* 1/2. Note that the diagonal gets mapped into the x-axis after a half
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* flip.
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*
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* How Many Bits Do We Need, Anyway?
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* ---------------------------------
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*
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* Note that if the line is visible on your screen, you must light up
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* exactly the correct pixels, no matter where in the 28.4 x 28.4 device
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* space the end points of the line lie (meaning you must handle 32 bit
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* DDAs, you can certainly have optimized cases for lesser DDAs).
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*
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* We move the origin to (floor(M0 / F), floor(N0 / F)), so when we
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* calculate gamma from (5), we know that 0 <= M0, N0 < F. And we
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* are in the first octant, so dM >= dN. Then we know that gamma can
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* be in the range [(-1/2)dM, (3/2)dM]. The DDI guarantees us that
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* valid lines will have dM and dN values at most 31 bits (unsigned)
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* of significance. So gamma requires 33 bits of significance (we store
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* this as a 64 bit number for convenience).
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*
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* When running through the DDA loop, r + dR can have a value in the
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* j
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* range 0 <= r < 2 dN; thus the result must be a 32 bit unsigned value.
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* j
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*
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* History:
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* 4-May-1992 -by- J. Andrew Goossen [andrewgo]
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* Wrote it.
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\**************************************************************************/
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BOOL bLines(
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BMINFO* pbmi, // Info about pixel format
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POINTFIX* pptfxFirst, // Start of first line
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POINTFIX* pptfxBuf, // Pointer to buffer of all remaining lines
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RUN* prun, // Pointer to runs if doing complex clipping
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ULONG cptfx, // Count of points in pptfxBuf or runs in prun
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LINESTATE* pls, // Color and style info
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RECTL* prclClip, // Clip rectangle if doing simple clipping
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PFNSTRIP *apfn, // Array of strip functions
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FLONG flStart, // Flags for each line
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CHUNK* pchBits, // Pointer to bitmap
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LONG lDelta) // Offset between rows of pixels (in CHUNKs)
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{
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POINTFIX* pptfxBufEnd = pptfxBuf + cptfx; // Last point in path record
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STYLEPOS spThis;
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ULONG M0;
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ULONG dM;
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ULONG N0;
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ULONG dN;
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ULONG dN_Original;
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ULONG N1;
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ULONG M1;
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FLONG fl;
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LONG x;
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LONG y;
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LONGLONG eqGamma;
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LONGLONG eqBeta;
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ULONG ulDelta;
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ULONG x0;
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ULONG y0;
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ULONG x1;
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ULONG x0_Original;
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ULONG y0_Original;
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ULONG x1_Original;
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POINTL ptlStart;
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STRIP strip;
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PFNSTRIP pfn;
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LONG* plStripEnd;
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ULONGLONG euq; // Temporary unsigned double-dword variable
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LONGLONG eq; // Temporary signed double-dword variable
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do {
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/***********************************************************************\
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* Start the DDA calculations. *
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\***********************************************************************/
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M0 = (LONG) pptfxFirst->x;
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dM = (LONG) pptfxBuf->x;
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N0 = (LONG) pptfxFirst->y;
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dN = (LONG) pptfxBuf->y;
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fl = flStart;
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if ((LONG) dM < (LONG) M0)
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{
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// Ensure that we run left-to-right:
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register ULONG ulTmp;
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SWAPL(M0, dM, ulTmp);
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SWAPL(N0, dN, ulTmp);
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fl |= FL_FLIP_H;
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}
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|
|
// We now have a line running left-to-right from (M0, N0) to
|
|
// (M0 + dM, N0 + dN):
|
|
|
|
if ((LONG) dN < (LONG) N0)
|
|
{
|
|
// Line runs from bottom to top, so flip across y = 0:
|
|
|
|
N0 = -(LONG) N0;
|
|
dN = -(LONG) dN;
|
|
fl |= FL_FLIP_V;
|
|
}
|
|
|
|
// Compute the deltas. The DDI says we can never have a valid delta
|
|
// with a magnitude more than 2^31 - 1, but the engine never actually
|
|
// checks its transforms. To ensure that we'll never puke on our shoes,
|
|
// we check for that case and simply refuse to draw the line:
|
|
|
|
dM -= M0;
|
|
if ((LONG) dM < 0)
|
|
goto Next_Line;
|
|
|
|
dN -= N0;
|
|
if ((LONG) dN < 0)
|
|
goto Next_Line;
|
|
|
|
if (dN >= dM)
|
|
{
|
|
if (dN == dM)
|
|
{
|
|
// Have to special case slopes of one:
|
|
|
|
fl |= FL_FLIP_SLOPE_ONE;
|
|
}
|
|
else
|
|
{
|
|
// Since line has slope greater than 1, flip across x = y:
|
|
|
|
register ULONG ulTmp;
|
|
SWAPL(dM, dN, ulTmp);
|
|
SWAPL(M0, N0, ulTmp);
|
|
fl |= FL_FLIP_D;
|
|
}
|
|
}
|
|
|
|
fl |= gaflRound[(fl & FL_ROUND_MASK) >> FL_ROUND_SHIFT];
|
|
|
|
x = LFLOOR((LONG) M0);
|
|
y = LFLOOR((LONG) N0);
|
|
|
|
M0 = FXFRAC(M0);
|
|
N0 = FXFRAC(N0);
|
|
|
|
{
|
|
// Calculate the remainder term [ dM * (N0 + F/2) - M0 * dN ]:
|
|
|
|
eqGamma = Int32x32To64((LONG) dM, N0 + F/2);
|
|
eq = Int32x32To64(M0, (LONG) dN);
|
|
|
|
eqGamma -= eq;
|
|
|
|
if (fl & FL_V_ROUND_DOWN)
|
|
eqGamma -= 1L; // Adjust so y = 1/2 rounds down
|
|
|
|
eqGamma >>= FLOG2;
|
|
|
|
eqBeta = ~eqGamma;
|
|
}
|
|
|
|
/***********************************************************************\
|
|
* Figure out which pixels are at the ends of the line. *
|
|
\***********************************************************************/
|
|
|
|
// Calculate x0, x1:
|
|
|
|
N1 = FXFRAC(N0 + dN);
|
|
M1 = FXFRAC(M0 + dM);
|
|
|
|
x1 = LFLOOR(M0 + dM);
|
|
|
|
// The toughest part of GIQ is determining the start and end pels.
|
|
//
|
|
// Our approach here is to calculate x0 and x1 (the inclusive start
|
|
// and end columns of the line respectively, relative to our normalized
|
|
// origin). Then x1 - x0 + 1 is the number of pels in the line. The
|
|
// start point is easily calculated by plugging x0 into our line equation
|
|
// (which takes care of whether y = 1/2 rounds up or down in value)
|
|
// getting y0, and then undoing the normalizing flips to get back
|
|
// into device space.
|
|
//
|
|
// We look at the fractional parts of the coordinates of the start and
|
|
// end points, and call them (M0, N0) and (M1, N1) respectively, where
|
|
// 0 <= M0, N0, M1, N1 < 16. We plot (M0, N0) on the following grid
|
|
// to determine x0:
|
|
//
|
|
// +-----------------------> +x
|
|
// |
|
|
// | 0 1
|
|
// | 0123456789abcdef
|
|
// |
|
|
// | 0 ........?xxxxxxx
|
|
// | 1 ..........xxxxxx
|
|
// | 2 ...........xxxxx
|
|
// | 3 ............xxxx
|
|
// | 4 .............xxx
|
|
// | 5 ..............xx
|
|
// | 6 ...............x
|
|
// | 7 ................
|
|
// | 8 ................
|
|
// | 9 ......**........
|
|
// | a ........****...x
|
|
// | b ............****
|
|
// | c .............xxx****
|
|
// | d ............xxxx ****
|
|
// | e ...........xxxxx ****
|
|
// | f ..........xxxxxx
|
|
// |
|
|
// | 2 3
|
|
// v
|
|
//
|
|
// +y
|
|
//
|
|
// This grid accounts for the appropriate rounding of GIQ and last-pel
|
|
// exclusion. If (M0, N0) lands on an 'x', x0 = 2. If (M0, N0) lands
|
|
// on a '.', x0 = 1. If (M0, N0) lands on a '?', x0 rounds up or down,
|
|
// depending on what flips have been done to normalize the line.
|
|
//
|
|
// For the end point, if (M1, N1) lands on an 'x', x1 =
|
|
// floor((M0 + dM) / 16) + 1. If (M1, N1) lands on a '.', x1 =
|
|
// floor((M0 + dM)). If (M1, N1) lands on a '?', x1 rounds up or down,
|
|
// depending on what flips have been done to normalize the line.
|
|
//
|
|
// Lines of exactly slope one require a special case for both the start
|
|
// and end. For example, if the line ends such that (M1, N1) is (9, 1),
|
|
// the line has gone exactly through (8, 0) -- which may be considered
|
|
// to be part of 'x' because of rounding! So slopes of exactly slope
|
|
// one going through (8, 0) must also be considered as belonging in 'x'.
|
|
//
|
|
// For lines that go left-to-right, we have the following grid:
|
|
//
|
|
// +-----------------------> +x
|
|
// |
|
|
// | 0 1
|
|
// | 0123456789abcdef
|
|
// |
|
|
// | 0 xxxxxxxx?.......
|
|
// | 1 xxxxxxx.........
|
|
// | 2 xxxxxx..........
|
|
// | 3 xxxxx...........
|
|
// | 4 xxxx............
|
|
// | 5 xxx.............
|
|
// | 6 xx..............
|
|
// | 7 x...............
|
|
// | 8 x...............
|
|
// | 9 x.....**........
|
|
// | a xx......****....
|
|
// | b xxx.........****
|
|
// | c xxxx............****
|
|
// | d xxxxx........... ****
|
|
// | e xxxxxx.......... ****
|
|
// | f xxxxxxx.........
|
|
// |
|
|
// | 2 3
|
|
// v
|
|
//
|
|
// +y
|
|
//
|
|
// This grid accounts for the appropriate rounding of GIQ and last-pel
|
|
// exclusion. If (M0, N0) lands on an 'x', x0 = 0. If (M0, N0) lands
|
|
// on a '.', x0 = 1. If (M0, N0) lands on a '?', x0 rounds up or down,
|
|
// depending on what flips have been done to normalize the line.
|
|
//
|
|
// For the end point, if (M1, N1) lands on an 'x', x1 =
|
|
// floor((M0 + dM) / 16) - 1. If (M1, N1) lands on a '.', x1 =
|
|
// floor((M0 + dM)). If (M1, N1) lands on a '?', x1 rounds up or down,
|
|
// depending on what flips have been done to normalize the line.
|
|
//
|
|
// Lines of exactly slope one must be handled similarly to the right-to-
|
|
// left case.
|
|
|
|
if (fl & FL_FLIP_H)
|
|
{
|
|
// ---------------------------------------------------------------
|
|
// Line runs right-to-left: <----
|
|
|
|
// Compute x1:
|
|
|
|
if (N1 == 0)
|
|
{
|
|
if (LROUND(M1, fl & FL_H_ROUND_DOWN))
|
|
{
|
|
x1++;
|
|
}
|
|
}
|
|
else if (ABS((LONG) (N1 - F/2)) + M1 > F)
|
|
{
|
|
x1++;
|
|
}
|
|
|
|
if ((fl & (FL_FLIP_SLOPE_ONE | FL_H_ROUND_DOWN))
|
|
== (FL_FLIP_SLOPE_ONE))
|
|
{
|
|
// Have to special-case diagonal lines going through our
|
|
// the point exactly equidistant between two horizontal
|
|
// pixels, if we're supposed to round x=1/2 down:
|
|
|
|
if ((N1 > 0) && (M1 == N1 + 8))
|
|
x1++;
|
|
|
|
// Don't you love special cases? Is this a rhetorical question?
|
|
|
|
if ((N0 > 0) && (M0 == N0 + 8))
|
|
{
|
|
x0 = 2;
|
|
ulDelta = dN;
|
|
goto right_to_left_compute_y0;
|
|
}
|
|
}
|
|
|
|
// Compute x0:
|
|
|
|
x0 = 1;
|
|
ulDelta = 0;
|
|
if (N0 == 0)
|
|
{
|
|
if (LROUND(M0, fl & FL_H_ROUND_DOWN))
|
|
{
|
|
x0 = 2;
|
|
ulDelta = dN;
|
|
}
|
|
}
|
|
else if (ABS((LONG) (N0 - F/2)) + M0 > F)
|
|
{
|
|
x0 = 2;
|
|
ulDelta = dN;
|
|
}
|
|
|
|
// Compute y0:
|
|
|
|
right_to_left_compute_y0:
|
|
|
|
y0 = 0;
|
|
eq = eqGamma;
|
|
eq += (LONGLONG) (ULONGLONG) ulDelta;
|
|
if ((eq >> 32) >= 0)
|
|
{
|
|
if ((eq >> 32) > 0 || (ULONG) eq >= 2 * dM - dN)
|
|
y0 = 2;
|
|
else if ((ULONG) eq >= dM - dN)
|
|
y0 = 1;
|
|
}
|
|
}
|
|
else
|
|
{
|
|
// ---------------------------------------------------------------
|
|
// Line runs left-to-right: ---->
|
|
//
|
|
|
|
// Compute x1:
|
|
|
|
x1--;
|
|
|
|
if (M1 > 0)
|
|
{
|
|
if (N1 == 0)
|
|
{
|
|
if (LROUND(M1, fl & FL_H_ROUND_DOWN))
|
|
x1++;
|
|
}
|
|
else if (ABS((LONG) (N1 - F/2)) <= (LONG) M1)
|
|
{
|
|
x1++;
|
|
}
|
|
}
|
|
|
|
if ((fl & (FL_FLIP_SLOPE_ONE | FL_H_ROUND_DOWN))
|
|
== (FL_FLIP_SLOPE_ONE | FL_H_ROUND_DOWN))
|
|
{
|
|
// Have to special-case diagonal lines going through our
|
|
// the point exactly equidistant between two horizontal
|
|
// pixels, if we're supposed to round x=1/2 down:
|
|
|
|
if ((M1 > 0) && (N1 == M1 + 8))
|
|
x1--;
|
|
|
|
if ((M0 > 0) && (N0 == M0 + 8))
|
|
{
|
|
x0 = 0;
|
|
goto left_to_right_compute_y0;
|
|
}
|
|
}
|
|
|
|
// Compute x0:
|
|
|
|
x0 = 0;
|
|
if (M0 > 0)
|
|
{
|
|
if (N0 == 0)
|
|
{
|
|
if (LROUND(M0, fl & FL_H_ROUND_DOWN))
|
|
x0 = 1;
|
|
}
|
|
else if (ABS((LONG) (N0 - F/2)) <= (LONG) M0)
|
|
{
|
|
x0 = 1;
|
|
}
|
|
}
|
|
|
|
// Compute y0:
|
|
|
|
left_to_right_compute_y0:
|
|
|
|
y0 = 0;
|
|
if ((eqGamma>>32) >= 0 &&
|
|
(ULONG) eqGamma >= dM - (dN & (-(LONG) x0)))
|
|
{
|
|
y0 = 1;
|
|
}
|
|
}
|
|
|
|
y0_Original = y0;
|
|
x0_Original = x0;
|
|
x1_Original = x1;
|
|
|
|
if ((LONG) x1 < (LONG) x0)
|
|
goto Next_Line;
|
|
|
|
/***********************************************************************\
|
|
* Complex Clipping. *
|
|
\***********************************************************************/
|
|
|
|
if (fl & FL_COMPLEX_CLIP)
|
|
{
|
|
dN_Original = dN;
|
|
|
|
Continue_Complex_Clipping:
|
|
|
|
if (fl & FL_FLIP_H)
|
|
{
|
|
// Line runs right-to-left <-----
|
|
|
|
x0 = x1_Original - prun->iStop;
|
|
x1 = x1_Original - prun->iStart;
|
|
}
|
|
else
|
|
{
|
|
// Line runs left-to-right ----->
|
|
|
|
x0 = x0_Original + prun->iStart;
|
|
x1 = x0_Original + prun->iStop;
|
|
}
|
|
|
|
prun++;
|
|
|
|
// Reset some variables we'll nuke a little later:
|
|
|
|
dN = dN_Original;
|
|
pls->spNext = pls->spComplex;
|
|
|
|
euq = UInt32x32To64(x0, dN);
|
|
euq += eqGamma;
|
|
|
|
y0 = DIV(euq,dM);
|
|
}
|
|
|
|
/***********************************************************************\
|
|
* Style calculations. *
|
|
\***********************************************************************/
|
|
|
|
if (fl & FL_STYLED)
|
|
{
|
|
ULONG cpelStyleLine; // # of style pels in entire line
|
|
ULONG cpelStyleFromThis; // # of style pels from first pel of line
|
|
STYLEPOS sp; // Style state at x0
|
|
|
|
// For those rare devices where xStep != yStep, for complex clipped
|
|
// lines that are non-major styled and have multiple runs, we will
|
|
// be doing more work than we have to:
|
|
|
|
BOOL bOffStyled = FALSE;
|
|
|
|
ULONG xStep = pls->xStep;
|
|
ULONG yStep = pls->yStep;
|
|
|
|
spThis = pls->spNext;
|
|
|
|
if (fl & FL_FLIP_D)
|
|
{
|
|
register ULONG ulTmp;
|
|
SWAPL(xStep, yStep, ulTmp);
|
|
}
|
|
|
|
if (xStep != yStep)
|
|
{
|
|
bOffStyled = UInt32x32To64(yStep, dN) > UInt32x32To64(xStep, dM);
|
|
}
|
|
|
|
// xStep is now the major style direction step size.
|
|
|
|
if (bOffStyled)
|
|
{
|
|
ULONG y1_Original;
|
|
|
|
// We need the original y1, so we simply compute it:
|
|
|
|
euq = UInt32x32To64(x1_Original, dN);
|
|
euq += eqGamma;
|
|
y1_Original = DIV(euq,dM);
|
|
|
|
// Our line is x-major but y-styled, or y-major and x-styled.
|
|
// We use xStep as the style-major step size, so adjust it:
|
|
|
|
xStep = yStep;
|
|
|
|
pls->ulStepRun = 0;
|
|
pls->ulStepSide = xStep;
|
|
pls->ulStepDiag = xStep;
|
|
|
|
cpelStyleLine = (y1_Original - y0_Original + 1);
|
|
|
|
if (fl & FL_FLIP_H)
|
|
cpelStyleFromThis = (y1_Original - y0 + 1);
|
|
else
|
|
cpelStyleFromThis = (y0 - y0_Original);
|
|
}
|
|
else
|
|
{
|
|
// Our line is x-major and x-styled, or y-major and y-styled:
|
|
|
|
pls->ulStepRun = xStep;
|
|
pls->ulStepSide = 0;
|
|
pls->ulStepDiag = xStep;
|
|
|
|
cpelStyleLine = (x1_Original - x0_Original + 1);
|
|
|
|
if (fl & FL_FLIP_H)
|
|
cpelStyleFromThis = (x1_Original - x0 + 1);
|
|
else
|
|
cpelStyleFromThis = (x0 - x0_Original);
|
|
}
|
|
|
|
// Note: we may overflow here if step size is more than 15:
|
|
|
|
sp = pls->spNext + xStep * cpelStyleFromThis;
|
|
pls->spNext = pls->spNext + xStep * cpelStyleLine;
|
|
|
|
// Normalize (being sure to cast to unsigned values because
|
|
// we want unsigned divides and positive results even if we
|
|
// overflowed):
|
|
|
|
if ((ULONG) sp >= (ULONG) pls->spTotal2)
|
|
sp = (ULONG) sp % pls->spTotal2;
|
|
|
|
if ((ULONG) pls->spNext >= (ULONG) pls->spTotal2)
|
|
pls->spNext = (ULONG) pls->spNext % pls->spTotal2;
|
|
|
|
// Since we always draw the line left-to-right, but styling is
|
|
// always done in the direction of the original style, we have
|
|
// to figure out where we are in the style array for the left
|
|
// edge of this line:
|
|
|
|
if (fl & FL_FLIP_H)
|
|
{
|
|
// Line originally ran right-to-left:
|
|
|
|
sp = -sp;
|
|
if (sp < 0)
|
|
sp += pls->spTotal2;
|
|
|
|
pls->bIsGap = !pls->bStartGap;
|
|
pls->pspStart = &pls->aspRightToLeft[0];
|
|
pls->pspEnd = &pls->aspRightToLeft[pls->cStyle - 1];
|
|
}
|
|
else
|
|
{
|
|
pls->bIsGap = pls->bStartGap;
|
|
pls->pspStart = &pls->aspLeftToRight[0];
|
|
pls->pspEnd = &pls->aspLeftToRight[pls->cStyle - 1];
|
|
}
|
|
|
|
// If the style array is of odd length, and we are on the second
|
|
// (or fourth, or sixth...) pass of the style array, the sense
|
|
// of the current array value is reversed (gap -> dash or dash
|
|
// -> gap):
|
|
|
|
if (sp >= pls->spTotal)
|
|
{
|
|
sp -= pls->spTotal;
|
|
if (pls->cStyle & 1)
|
|
pls->bIsGap = !pls->bIsGap;
|
|
}
|
|
|
|
// Find our position in the style array:
|
|
|
|
pls->psp = pls->pspStart;
|
|
while (sp >= *pls->psp)
|
|
sp -= *pls->psp++;
|
|
|
|
ASSERTGDI(pls->psp <= pls->pspEnd, "Flew into NeverNeverLand");
|
|
|
|
pls->spRemaining = *pls->psp - sp;
|
|
|
|
// Our position in the style array tells us if we're currently
|
|
// working on a gap or a dash:
|
|
|
|
if ((pls->psp - pls->pspStart) & 1)
|
|
pls->bIsGap = !pls->bIsGap;
|
|
}
|
|
|
|
// Flip back to device space:
|
|
|
|
ptlStart.x = x + x0;
|
|
ptlStart.y = y + y0;
|
|
|
|
if (fl & FL_FLIP_D)
|
|
{
|
|
register LONG lTmp;
|
|
SWAPL(ptlStart.x, ptlStart.y, lTmp);
|
|
}
|
|
|
|
if (fl & FL_FLIP_V)
|
|
{
|
|
ptlStart.y = -ptlStart.y;
|
|
}
|
|
|
|
if (2 * dN > dM)
|
|
{
|
|
// Do a half flip!
|
|
|
|
fl |= FL_FLIP_HALF;
|
|
|
|
eqBeta = eqGamma;
|
|
eqBeta -= (LONGLONG) (ULONGLONG) dM;
|
|
|
|
dN = dM - dN;
|
|
y0 = x0 - y0; // Note this may overflow, but that's okay
|
|
}
|
|
|
|
// Now, run the DDA starting at (ptlStart.x, ptlStart.y)!
|
|
|
|
strip.flFlips = fl;
|
|
pfn = apfn[(fl & FL_STRIP_MASK) >> FL_STRIP_SHIFT];
|
|
|
|
strip.iPixel = ptlStart.x & pbmi->maskPixel;
|
|
strip.lDelta = lDelta;
|
|
strip.pchScreen = pchBits + ptlStart.y * lDelta;
|
|
|
|
if (pbmi->cShift < 0)
|
|
{
|
|
// 24bpp takes the address of the first byte:
|
|
|
|
strip.pchScreen = (CHUNK*) ((BYTE*) strip.pchScreen + ptlStart.x * 3);
|
|
}
|
|
else
|
|
{
|
|
// All other formats take the address of the first dword:
|
|
|
|
strip.pchScreen += (ptlStart.x >> pbmi->cShift);
|
|
}
|
|
|
|
plStripEnd = &strip.alStrips[STRIP_MAX];
|
|
|
|
// Now calculate the DDA variables needed to figure out how many pixels
|
|
// go in the very first strip:
|
|
|
|
{
|
|
register LONG* plStrip = &strip.alStrips[0];
|
|
register LONG cPels = x1 - x0 + 1;
|
|
register LONG i;
|
|
register ULONG r;
|
|
register ULONG dI;
|
|
register ULONG dR;
|
|
|
|
if (dN == 0)
|
|
i = LONG_MAX;
|
|
else
|
|
{
|
|
euq = UInt32x32To64(y0 + 1, dM);
|
|
euq += eqBeta;
|
|
|
|
i = DIVREM(euq, dN, &r) - x0 + 1;
|
|
|
|
dI = dM / dN;
|
|
dR = dM % dN; // 0 <= dR < dN
|
|
}
|
|
|
|
ASSERTGDI(i > 0 && i <= LONG_MAX, "Weird start strip length");
|
|
ASSERTGDI(cPels > 0, "Zero cPels");
|
|
|
|
/***********************************************************************\
|
|
* Run the DDA! *
|
|
\***********************************************************************/
|
|
|
|
while(TRUE)
|
|
{
|
|
cPels -= i;
|
|
if (cPels <= 0)
|
|
break;
|
|
|
|
*plStrip++ = i;
|
|
|
|
if (plStrip == plStripEnd)
|
|
{
|
|
//Sundown: safe to truncate to LONG since will not exceed cPels
|
|
strip.cStrips = (LONG)(plStrip - &strip.alStrips[0]);
|
|
(*pfn)(&strip, pbmi, pls);
|
|
plStrip = &strip.alStrips[0];
|
|
}
|
|
|
|
i = dI;
|
|
r += dR;
|
|
|
|
if (r >= dN)
|
|
{
|
|
r -= dN;
|
|
i++;
|
|
}
|
|
}
|
|
|
|
*plStrip++ = cPels + i;
|
|
|
|
//Sundown safe truncation.
|
|
strip.cStrips = (LONG)(plStrip - &strip.alStrips[0]);
|
|
(*pfn)(&strip, pbmi, pls);
|
|
}
|
|
|
|
Next_Line:
|
|
|
|
// We're done with that run. Figure out what to do next:
|
|
|
|
if (fl & FL_COMPLEX_CLIP)
|
|
{
|
|
cptfx--;
|
|
if (cptfx != 0)
|
|
goto Continue_Complex_Clipping;
|
|
|
|
break;
|
|
}
|
|
else
|
|
{
|
|
pptfxFirst = pptfxBuf;
|
|
pptfxBuf++;
|
|
}
|
|
|
|
} while (pptfxBuf < pptfxBufEnd);
|
|
|
|
return(TRUE);
|
|
}
|