运用

Stopwatch();

我发现循环的内部每个循环需要.000000000022秒.

每个循环完成循环的外部需要.0000077秒.

我发现从开始到图像保存的整个功能每次运行运行.21秒.

从内部循环计算完成图像：

.000000000022 x 640 = .000000001408 seconds
.000000001408 x 360 = .00000050688  seconds

从外部循环计算到完成图像：

.0000077 x 360 = .002772 seconds

如果我可以创建一个与我将被设置的时间相关的图像,但运行整个代码的代码需要.21秒来完成所有代码

temp_byte1 = main_byte1;
temp_byte2 = main_byte2;

timer1.Reset();
timer1.Start();

Bitmap mybmp = new Bitmap(1280,720);
BitmapData BPD = mybmp.LockBits(new Rectangle(0,1280,720),ImageLockMode.WriteOnly,mybmp.PixelFormat);
IntPtr xptr = BPD.Scan0;
IntPtr yptr = BPD.Scan0;
yptr = new IntPtr( yptr.ToInt64() + (1280 * 720 * 2));
int bytes = Math.Abs(BPD.Stride);
byte[][] rgb = new byte[720][];
int Y1,Y2,Y3,Y4,Y5,Y6,Y7,Y8;
int U1,U2,V1,V2,U3,U4,V3,V4;
for (int one = 0; one < 360; one++)
{
    timer2.Reset();
    timer2.Start();
    rgb[one] = new byte[bytes];
    rgb[360 + one] = new byte[bytes];
    for (int two = 0; two < 640; two++)
    {
        timer3.Reset();
        timer3.Start();
        U1 = temp_byte1[one * 2560 + 4 * two + 0];
        Y1 = temp_byte1[one * 2560 + 4 * two + 1];
        V1 = temp_byte1[one * 2560 + 4 * two + 2];
        Y2 = temp_byte1[one * 2560 + 4 * two + 3];

        U2 = temp_byte2[one * 2560 + 4 * two + 0];
        Y3 = temp_byte2[one * 2560 + 4 * two + 1];
        V2 = temp_byte2[one * 2560 + 4 * two + 2];
        Y4 = temp_byte2[one * 2560 + 4 * two + 3];

        RGB_Conversion(Y1,U1,two * 8 + 0,rgb[one]);
        RGB_Conversion(Y2,two * 8 + 4,rgb[one]);

        RGB_Conversion(Y3,rgb[(360 + one)]);
        RGB_Conversion(Y4,rgb[(360 + one)]);

        timer3.Stop();
        timer3_[two] = timer3.Elapsed;
    }
    Marshal.Copy(rgb[one],xptr,5120);
    xptr = new IntPtr(xptr.ToInt64() + 5120);
    Marshal.Copy(rgb[(360 + one)],yptr,5120);
    yptr = new IntPtr(yptr.ToInt64() + 5120);
    timer2.Stop();
    timer2_[one] = timer2.Elapsed;
}
mybmp.UnlockBits(BPD);
mybmp.Save(GetDateTimeString("IP Pictures") + ".jpg",ImageFormat.Jpeg);

代码工作,它将yuv422传入的字节数组转换为完整大小的jpeg,但无法理解为什么for循环的速度和整个代码之间存在这样的差异

我感动了

byte[][]rgb = new byte[720];  
rgb[x] = new byte[bytes];

在程序启动时获取init而不是每个函数调用/运行没有可测量的速度增加的全局.

UPDATE

RGB转换：接收YUV并将其转换为RGB并将其放入保存值的全局数组中

public void RGB_Conversion(int Y,int U,int V,int MULT,byte[] rgb)
{

    int C,D,E;
    int R,G,B;

    // create the params for rgb conversion
    C = Y - 16;
    D = U - 128;
    E = V - 128;

    //R = clamp((298 x C + 409 x E + 128)>>8)
    //G = clamp((298 x C - 100 x D - 208 x E + 128)>>8)
    //B = clamp((298 x C + 516 x D + 128)>>8)

    R = (298 * C + 409 * E + 128)/256;
    G = (298 * C - 100 * D - 208 * E + 128)/256;
    B = (298 * C + 516 * D + 128)/256;

    if (R > 255)
        R = 255;
    if (R < 0)
        R = 0;
    if (G > 255)
        G = 255;
    if (G < 0)
        G = 0;
    if (B > 255)
        B = 255;
    if (B < 0)
        B = 0;

    rgb[MULT + 3] = 255;
    rgb[MULT + 0] = (byte)B;
    rgb[MULT + 1] = (byte)G;
    rgb[MULT + 2] = (byte)R;
    }