/// <summary>
/// Orders an array of three ResultPoints in an order [A,B,C] such that AB is less than AC and
/// BC is less than AC and the angle between BC and BA is less than 180 degrees.
/// </summary>
/// <param name="patterns">array of three <see cref="ResultPoint" /> to order</param>
public static void OrderBestPatterns(ResultPoint[] patterns)
{
// Find distances between pattern centers
float zeroOneDistance = Distance(patterns[0], patterns[1]);
float oneTwoDistance = Distance(patterns[1], patterns[2]);
float zeroTwoDistance = Distance(patterns[0], patterns[2]);
ResultPoint pointA, pointB, pointC;
// Assume one closest to other two is B; A and C will just be guesses at first
if (oneTwoDistance >= zeroOneDistance && oneTwoDistance >= zeroTwoDistance)
{
pointB = patterns[0];
pointA = patterns[1];
pointC = patterns[2];
}
else if (zeroTwoDistance >= oneTwoDistance && zeroTwoDistance >= zeroOneDistance)
{
pointB = patterns[1];
pointA = patterns[0];
pointC = patterns[2];
}
else
{
pointB = patterns[2];
pointA = patterns[0];
pointC = patterns[1];
}
// Use cross product to figure out whether A and C are correct or flipped.
// This asks whether BC x BA has a positive z component, which is the arrangement
// we want for A, B, C. If it's negative, then we've got it flipped around and
// should swap A and C.
if (CrossProductZ(pointA, pointB, pointC) < 0.0f)
{
ResultPoint temp = pointA;
pointA = pointC;
pointC = temp;
}
patterns[0] = pointA;
patterns[1] = pointB;
patterns[2] = pointC;
}
/// <summary>
/// calculates the distance between two points
/// </summary>
/// <param name="pattern1">first pattern</param>
/// <param name="pattern2">second pattern</param>
/// <returns>
/// distance between two points
/// </returns>
public static float Distance(ResultPoint pattern1, ResultPoint pattern2)
{
return(MathUtils.Distance(pattern1.X, pattern1.Y, pattern2.X, pattern2.Y));
}
public Result Decode(BinaryBitmap image)
{
int width = image.Width;
int height = image.Height;
BitArray row = new BitArray(width);
int rowStep = Math.Max(1, height >> 5);
int maxLines = 15; // 15 rows spaced 1/32 apart is roughly the middle half of the image
int middle = height >> 1;
for (int x = 0; x < maxLines; x++)
{
// Scanning from the middle out. Determine which row we're looking at next:
int rowStepsAboveOrBelow = (x + 1) >> 1;
bool isAbove = (x & 0x01) == 0; // i.e. is x even?
int rowNumber = middle + rowStep * (isAbove ? rowStepsAboveOrBelow : -rowStepsAboveOrBelow);
if (rowNumber < 0 || rowNumber >= height)
{
// Oops, if we run off the top or bottom, stop
break;
}
// Estimate black point for this row and load it:
row = image.GetBlackRow(rowNumber, row);
if (row == null)
{
continue;
}
// While we have the image data in a BitArray, it's fairly cheap to reverse it in place to
// handle decoding upside down barcodes.
for (int attempt = 0; attempt < 2; attempt++)
{
if (attempt == 1)
{
// trying again?
row.Reverse(); // reverse the row and continue
// This means we will only ever draw result points *once* in the life of this method
// since we want to avoid drawing the wrong points after flipping the row, and,
// don't want to clutter with noise from every single row scan -- just the scans
// that start on the center line.
}
// Look for a barcode
Result result = Reader.DecodeRow(rowNumber, row);
if (result == null)
{
continue;
}
// We found our barcode
if (attempt == 1)
{
// And remember to flip the result points horizontally.
ResultPoint[] points = result.ResultPoints;
if (points != null)
{
points[0] = new ResultPoint(width - points[0].X - 1, points[0].Y);
points[1] = new ResultPoint(width - points[1].X - 1, points[1].Y);
}
}
return(result);
}
}
return(null);
}