任何可以解释的BitmapData步幅的功能？ [英] any can explain the function of stride in bitmapdata?

问题描述

        Bitmap bit1 = new Bitmap( bmpimg , width , height );
        Bitmap bit2 = new Bitmap( bmp , width , height );

        Bitmap bmpresult = new Bitmap( width , height );

        BitmapData data1 = bit1.LockBits( new Rectangle( 0 , 0 , bit1.Width , bit1.Height ) , ImageLockMode.ReadWrite , PixelFormat.Format24bppRgb );
        BitmapData data2 = bit2.LockBits( new Rectangle( 0 , 0 , bit2.Width , bit2.Height ) , ImageLockMode.ReadWrite , PixelFormat.Format24bppRgb );
        BitmapData data3 = bmpresult.LockBits( new Rectangle( 0 , 0 , bmpresult.Width , bmpresult.Height ) , ImageLockMode.ReadWrite , PixelFormat.Format24bppRgb );

unsafe
        {
            int remain1 = data1.Stride - data1.Width * 3;
            int remain2 = data2.Stride - data2.Width * 3;
            int remain3 = data3.Stride - data3.Width * 3;


            byte* ptr1 = ( byte* )data1.Scan0;
            byte* ptr2 = ( byte* )data2.Scan0;
            byte* ptr3 = ( byte* )data3.Scan0;

            for( int i = 0 ; i < height ; i ++ )
            {
                for( int j = 0 ; j < width * 3 ; j ++ )
                {
                    ptr3[ 0 ] = ( byte ) ( XOR_Operator( ptr1[ 0 ] , ptr2[ 0 ] ) );
                    ptr1 ++;
                    ptr2 ++;
                    ptr3 ++;
                }

                ptr1 += remain1;
                ptr2 += remain2;
                ptr3 += remain3;
            }


        }

        bit1.UnlockBits( data1 );
        bit2.UnlockBits( data2 );
        bmpresult.UnlockBits( data3 );

        return bmpresult;
    }

是有必要维持获取数据对象。

is it necessary to get remain for data objects

推荐答案

步幅是存在的，因为已经不幸通过对API层泄露的硬件要求。

Stride is there due to hardware requirements that have unfortunately leaked through to the API layer.

这是重要的，因为Windows驱动程序，有时需要扫描线（图中的行）是内存对齐。这就是为什么他们有时会大于他们将严格需要的人。

It is important because Windows drivers sometimes require that scanlines (rows in the image) be memory aligned. That is why they are sometimes larger than they would strictly need to be.

例如见这个MSDN文章的DIB和其使用。

See for example this MSDN article DIBs and Their Use.

每个扫描线的 DWORD -aligned。该扫描线被缓冲到对齐;缓冲不一定0。

Every scanline is DWORD-aligned. The scanline is buffered to alignment; the buffering is not necessarily 0.

您处理似乎是足够的。

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