怎样旋转一个画布中的某个矩形Rect(T,L,R,B)区域,求救!
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bmp_rotate(Image1.Picture.Bitmap, Image2.Picture.Bitmap, RAngle);procedure TfrmColor.bmp_rotate(src,dst:tbitmap;angle:extended);
var
c1x,c1y,c2x,c2y:integer;
p1x,p1y,p2x,p2y:integer;
radius,n:integer;
alpha:extended;
c0,c1,c2,c3:tcolor;
begin
//将角度转换为PI值
angle := (angle / 180) * pi;
// 计算中心点,你可以修改它
c1x := src.width div 2;
c1y := src.height div 2;
c2x := dst.width div 2;
c2y := dst.height div 2; // 步骤数值number
if c2x < c2y then
n := c2y
else
n := c2x;
dec (n,1); // 开始旋转
for p2x := 0 to n do begin
for p2y := 0 to n do begin
if p2x = 0 then
alpha:= pi/2
else
alpha := arctan2(p2y,p2x);
radius := round(sqrt((p2x*p2x)+(p2y*p2y)));
p1x := round(radius * cos(angle+alpha));
p1y := round(radius * sin(angle+alpha));
c0 := src.canvas.pixels[c1x+p1x,c1y+p1y];
c1 := src.canvas.pixels[c1x-p1x,c1y-p1y];
c2 := src.canvas.pixels[c1x+p1y,c1y-p1x];
c3 := src.canvas.pixels[c1x-p1y,c1y+p1x]; dst.canvas.pixels[c2x+p2x,c2y+p2y]:=c0;
dst.canvas.pixels[c2x-p2x,c2y-p2y]:=c1;
dst.canvas.pixels[c2x+p2y,c2y-p2x]:=c2;
dst.canvas.pixels[c2x-p2y,c2y+p2x]:=c3;
end;
application.processmessages
end;
end;
*************8
----把一个点绕原点旋转α角度后,新的坐标位置与原坐标位置的
关系是:X=xcosα-ysinα
Y=xsinα+ycosα
例如要把位图顺时针旋转90度,坐标变换公式为:X=-yY=x----把这一公式用到Image构件上,显示位图的主要问题是Image构
件显示的位图只有一个象限,并且x、y坐标也是互相颠倒的,为了
解决这个问题,必须在Image构件上建立一个新的坐标原点。下面就
举例说明。----1.新建一工程project1,在form1上添加image1、image2、
image3、image4,其Autosize属性设为True,image1用来显示原
图,image2、image3、image4分别用来显示旋转90度、180度和270
度后的图像。双击image1,选定一幅bmp图。----2.添加Button1、Button2、Button3和Button4按钮,其
caption属性分别为"原图"、"旋转90度"、"旋转180度"、
"旋转270度"。----3.编写"旋转90度"按钮的OnClick事件。procedureTForm1.Button2Click(Sender:TObject);
var
i,j:integer;
begin
//确定旋转后位图的大小
image2.Picture.Bitmap.Height:=image1.picture.width;
image2.Picture.Bitmap.Width:=image1.picture.height;
fori:=0toimage1.Heightdo
forj:=0toimage1.Widthdo
image2.canvas.Pixels[(-i+image1.Height),
j]:=image1.canvas.Pixels[j,i];
end;
----4.编写"旋转180度"按钮的OnClick事件。procedureTForm1.Button3Click(Sender:TObject);
var
i,j:integer;
begin
//确定旋转后位图的大小
image3.Picture.Bitmap.Height:=image1.picture.Height;
image3.Picture.Bitmap.Width:=image1.picture.Width;
fori:=0toimage1.Heightdo
forj:=0toimage1.Widthdo
image3.canvas.Pixels[(image1.Width
-j),(image1.Height-i)]:=image1.canvas.Pixels[j,i];
end;----5.编写"旋转270度"按钮的OnClick事件。代码和步骤3相
似,只需要用image4替换image2,然后用以下的语句替换步骤3for
循环中的原有的语句。image4.canvas.Pixels[i,(image1.Width-j)]:=image1.canvas.Pixels[j,i];procedure Rotate(Bmp,Dst:TFastRGB;cx,cy:Integer;Angle:Extended);
var
cAngle,
sAngle: Double;
xDiff,
yDiff,
xpr,ypr,
ix,iy,
px,py,
x,y: Integer;
Tmp: PFColor;{what means?}
begin
Angle:=-Angle*Pi/180;
sAngle:=Sin(Angle);
cAngle:=Cos(Angle);
xDiff:=(Dst.Width-Bmp.Width)div 2;
yDiff:=(Dst.Height-Bmp.Height)div 2;
Tmp:=Dst.Bits;{what means?}
for y:=0 to Dst.Height-1 do
begin
py:=2*(y-cy)+1;
for x:=0 to Dst.Width-1 do
begin
px:=2*(x-cx)+1;
xpr:=Round(px*cAngle-py*sAngle);
ypr:=Round(px*sAngle+py*cAngle);
ix:=((xpr-1)div 2+cx)-xDiff;
iy:=((ypr-1)div 2+cy)-yDiff;
if(ix>-1)and(ix<Bmp.Width)and(iy>-1)and(iy<Bmp.Height)then
Tmp^:=Bmp.Pixels[iy,ix]; {what means?}
Inc(Tmp);
end;
Tmp:=Pointer(Integer(Tmp)+Dst.Gap); {what means?}
end;
end;原理:
cos(Alpha), sin(Alpha), 0
只需要用源矩阵乘以 -sin(Alpha),cos(Alpha), 0
0, 0, 1 如果你发现转过来的图形带有很整齐的花点,解决的办法是反向计算,即从目标求的源点的坐标和像素值。
以上的例子就是这样的。如果真的按下面下矩阵计算每个点,目标区有一些点会是白点(因为有些源点通过计算和四舍五入在目标中凑到一起了),我以前解决的办法是从目标求的源点的坐标和像素值,不过首先要取到目标区的区域(往往是斜的)。
cos(Alpha), sin(Alpha), 0
-sin(Alpha),cos(Alpha), 0
0, 0, 1 下载我说的控件吗,有现成的例子及DEMO!http://www.crosswinds.net/~khojasteh/delphi-components.htmlTRotateImage v1.21
This component is a visual component similar to TImage with ability to rotate the image in any arbitrary angle. TRotateImage can be used on Delphi 3, 4, and 5.
MaxPixelCount = 32768;TYPE
TRGBTripleArray = ARRAY[0..MaxPixelCount-1] OF TRGBTriple;
pRGBTripleArray = ^TRGBTripleArray;
...// "Simple" approach. For pixel (i,j), use "reverse" rotation to find
// where the rotated pixel must have been before the rotation.
// Don't bother with center of pixel adjustment.
// Assumes input BitmapOriginal has PixelFormat = pf24bit.
FUNCTION RotateBitmapMethod1 (CONST BitmapOriginal: TBitmap;
CONST iRotationAxis, jRotationAxis: INTEGER;
CONST AngleOfRotation: DOUBLE {radians} ): TBitmap; VAR
cosTheta : EXTENDED;
i : INTEGER;
iOriginal : INTEGER;
iPrime : INTEGER;
j : INTEGER;
jOriginal : INTEGER;
jPrime : INTEGER;
RowOriginal: pRGBTripleArray;
RowRotated : pRGBTRipleArray;
sinTheta : EXTENDED;
BEGIN
// The size of BitmapRotated is the same as BitmapOriginal. PixelFormat
// must also match since 24-bit GBR triplets are assumed in ScanLine.
RESULT := TBitmap.Create;
RESULT.Width := BitmapOriginal.Width;
RESULT.Height := BitmapOriginal.Height;
RESULT.PixelFormat := pf24bit; // Force this // Get SIN and COS in single call from math library
sincos(AngleOfRotation, sinTheta, cosTheta); // If no math library, then use this:
// sinTheta := SIN(AngleOfRotation);
// cosTheta := COS(AngleOfRotation); // Step through each row of rotated image.
FOR j := RESULT.Height-1 DOWNTO 0 DO
BEGIN
RowRotated := RESULT.Scanline[j];
jPrime := j - jRotationAxis; FOR i := RESULT.Width-1 DOWNTO 0 DO
BEGIN
iPrime := i - iRotationAxis;
iOriginal := iRotationAxis + ROUND(iPrime * CosTheta - jPrime * sinTheta);
jOriginal := jRotationAxis + ROUND(iPrime * sinTheta + jPrime * cosTheta); // Make sure (iOriginal, jOriginal) is in BitmapOriginal. If not,
// assign blue color to corner points.
IF (iOriginal >= 0) AND (iOriginal <= BitmapOriginal.Width-1) AND
(jOriginal >= 0) AND (jOriginal <= BitmapOriginal.Height-1)
THEN BEGIN
// Assign pixel from rotated space to current pixel in BitmapRotated
RowOriginal := BitmapOriginal.Scanline[jOriginal];
RowRotated[i] := RowOriginal[iOriginal]
END
ELSE BEGIN
RowRotated[i].rgbtBlue := 255; // assign "corner" color
RowRotated[i].rgbtGreen := 0;
RowRotated[i].rgbtRed := 0
END END
END
END {RotateBitmapMethod1};