本文整理匯總了C#中BinaryBitmap.rotateCounterClockwise方法的典型用法代碼示例。如果您正苦於以下問題：C# BinaryBitmap.rotateCounterClockwise方法的具體用法？C# BinaryBitmap.rotateCounterClockwise怎麽用？C# BinaryBitmap.rotateCounterClockwise使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類BinaryBitmap的用法示例。
在下文中一共展示了BinaryBitmap.rotateCounterClockwise方法的2個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的C#代碼示例。
示例1: decode
/// <summary>
/// Locates and decodes a barcode in some format within an image. This method also accepts
/// hints, each possibly associated to some data, which may help the implementation decode.
/// Note that we don't try rotation without the try harder flag, even if rotation was supported.
/// </summary>
/// <param name="image">image of barcode to decode</param>
/// <param name="hints">passed as a <see cref="IDictionary{TKey, TValue}"/> from <see cref="DecodeHintType"/>
/// to arbitrary data. The
/// meaning of the data depends upon the hint type. The implementation may or may not do
/// anything with these hints.</param>
/// <returns>
/// String which the barcode encodes
/// </returns>
virtual public Result decode(BinaryBitmap image, IDictionary<DecodeHintType, object> hints)
{
var result = doDecode(image, hints);
if (result == null)
{
bool tryHarder = hints != null && hints.ContainsKey(DecodeHintType.TRY_HARDER);
bool tryHarderWithoutRotation = hints != null && hints.ContainsKey(DecodeHintType.TRY_HARDER_WITHOUT_ROTATION);
if (tryHarder && !tryHarderWithoutRotation && image.RotateSupported)
{
BinaryBitmap rotatedImage = image.rotateCounterClockwise();
result = doDecode(rotatedImage, hints);
if (result == null)
return null;
// Record that we found it rotated 90 degrees CCW / 270 degrees CW
IDictionary<ResultMetadataType, object> metadata = result.ResultMetadata;
int orientation = 270;
if (metadata != null && metadata.ContainsKey(ResultMetadataType.ORIENTATION))
{
// But if we found it reversed in doDecode(), add in that result here:
orientation = (orientation +
(int) metadata[ResultMetadataType.ORIENTATION])%360;
}
result.putMetadata(ResultMetadataType.ORIENTATION, orientation);
// Update result points
ResultPoint[] points = result.ResultPoints;
if (points != null)
{
int height = rotatedImage.Height;
for (int i = 0; i < points.Length; i++)
{
points[i] = new ResultPoint(height - points[i].Y - 1, points[i].X);
}
}
}
}
return result;
}示例2: decode
// Note that we don't try rotation without the try harder flag, even if rotation was supported.
public virtual Result decode(BinaryBitmap image, System.Collections.Generic.Dictionary <Object,Object> hints)
{
try
{
return doDecode(image, hints);
}
catch (ReaderException re)
{
bool tryHarder = hints != null && hints.ContainsKey(DecodeHintType.TRY_HARDER);
if (tryHarder && image.RotateSupported)
{
BinaryBitmap rotatedImage = image.rotateCounterClockwise();
Result result = doDecode(rotatedImage, hints);
// Record that we found it rotated 90 degrees CCW / 270 degrees CW
System.Collections.Generic.Dictionary <Object,Object> metadata = result.ResultMetadata;
int orientation = 270;
if (metadata != null && metadata.ContainsKey(ResultMetadataType.ORIENTATION))
{
// But if we found it reversed in doDecode(), add in that result here:
orientation = (orientation + ((System.Int32) metadata[ResultMetadataType.ORIENTATION])) % 360;
}
result.putMetadata(ResultMetadataType.ORIENTATION, (System.Object) orientation);
// Update result points
ResultPoint[] points = result.ResultPoints;
int height = rotatedImage.Height;
for (int i = 0; i < points.Length; i++)
{
points[i] = new ResultPoint(height - points[i].Y - 1, points[i].X);
}
return result;
}
else
{
throw re;
}
}
}