关于YV16格式转换到I420格式的问题

Question

最近在弄YUV图像格式转换的问题，想要将YV16格式转换到I420，但是出来的结果是转换成了YV12，也就是转换结果中的UV分量互换了。
代码如下，想确认下YV16中的UV分量存储顺序是V前，U后吧，那这样的话代码逻辑应该是没有问题的？
另外我也看到过一个别人分享的YV16转I420，看他的代码逻辑，表现出来的YV16格式是U前，V后，所以就很困惑，还望有人能够解惑！

//----------------------------------我的代码
/** YV16 422P
yyyy yyyy
vvvv
uuuu

-> I420 420P    
yyyy yyyy
uu
vv
*/

void yuv422p_to_yuv420p(unsigned char* yuv422p, unsigned char* yuv420p, int width, int height)
{
 
    int i, j;
    int y_size = width * height;
 
    unsigned char* y = yuv422p;
    unsigned char* v = yuv422p + y_size;
    unsigned char* u = yuv422p + y_size + y_size/2;
 
    unsigned char* y_tmp = yuv420p;
    unsigned char* u_tmp = yuv420p + y_size;
    unsigned char* v_tmp = yuv420p + y_size + y_size/ 4;
 
    // y
    memcpy(y_tmp, y, y_size);
 
    // u
    for(j=0; j< height; j=j+2)
        {
            for(i=0; i<width/2; i=i++)
            {
                    u_tmp[width/2 * (j/2) + i] = u[width/2 * j + i];
                  v_tmp[width/2 * (j/2) + i] = v[width/2 * j + i];
                
            }
        }
}

//------------------别人的代码
int yv16_to_i420(unsigned char* pImageBuf, int width, int height) 
{ 
    int i, j; 
    int y_size,uvw;
    unsigned char* p_u; 
    unsigned char* p_v; 
   if(NULL == pImageBuf)
        return -1;
    uvw = width>>1;
    y_size = width * height; 
    p_u  = pImageBuf + y_size;
    p_v  = p_u + (y_size>>1);
    i = uvw;
 
    for(j = 2;j<height;j+=2)
    {
       memcpy(p_u + i,p_u+j*uvw, uvw);
       i+= uvw;
    }
    p_u += (y_size>>2);
    i = 0;
    for (j = 0; j < height; j+=2) 
    {
       memcpy(p_u + i, p_v+j*uvw,uvw);
       i+= uvw;
    }
    return 0;
}