Use iplRotate the picture what I get is losing the part of left and right .why?

Question

the code is create CreateImageHeader for two :imageSo and imageDe, then value imagedata by scanline , then rotate by judge angle, then value scanline by imagedata,

I show the main code like what I said as behind,hope someone could give me the suggestion or thinking.

create:

imageSo = CreateImageHeader(bmp->PixelFormat, bmp->Width,bmp->Height);  
imageDe = CreateImageHeader(bmp->PixelFormat, bmp->Width,bmp->Height);

shift:

bit1=(BYTE*) bmp->ScanLine[hSrc-1];  
bit2=imageSo->imageData;  
for(int i=0;i < hSrc;i++)    
{  
    memcpy(bit2,bit1,cpy);  
    bit1+=wBmp;  
    bit2+=wIpl;  
}

rotate:

if(param.Angle){  

double xshift=0;  

double yshift=0;  

iiplGetRotateShift(bmp->Width/2,bmp->Height/2,BmpProperty.Angle,&xshift,&yshift);    

iiplRotate(imageSo,imageDe,BmpProperty.Angle,xshift,yshift,IPL_INTER_CUBIC|IPL_SMOOTH_EDGE);  
}

shift2:

bit1=(BYTE*) bmp->ScanLine[hSrc-1];  
bit2=imageSo->imageData;  
for(int i=0;i < hSrc;i++)   
{  
    memcpy(bit1,bit2,cpy);  
    bit1+=wBmp;  
    bit2+=wIpl;  
}

I thought my approach is wrong,I printed the picture and found the black border is still here.sad.... I need to think about the black border again. — fading, Aug 16 '17 at 01:17

score 0 · Accepted Answer · answered Aug 30 '17 at 03:34

I know the reason now , because after rotated, the canvas was also rotating,so I set the canvas to be a square which was rectangle before.also understand the picture in canvas's position would also help.

here is the website which is introduce the details of canvas rotate. enter link description here

Use iplRotate the picture what I get is losing the part of left and right .why?

1 Answers1