2013-06-04 103 views
0

我创建了一个无分支中位数3x3过滤器,并且在高分辨率图像(约4K下3K)时获得了每通过200毫米左右的数据,我认为我可以获得如果我创建了内核来利用工作组,那就更好了。不幸的是,该工具告诉我我做错了什么,我想知道。OpenCL使用本地工作组的图像过滤器

#define wgs 16 
//Work group size 
#define cas3(a, b)        \ 
    do {           \ 
      float4 x = a;       \ 
      int4 c = a> b;       \ 
      a.s012 = select(b, a, c).s012;   \ 
      b.s012 = select(x, b, c).s012;   \ 
    } while (0) 

__kernel void median3x3_rgb(read_only image2d_t src, write_only image2d_t dst) { 
    int gx = get_global_id(0), gy = get_global_id(1); 
    int lx = get_local_id(0), ly = get_local_id(1); 
const sampler_t smp = CLK_NORMALIZED_COORDS_FALSE | CLK_ADDRESS_CLAMP_TO_EDGE | CLK_FILTER_NEAREST; 

    if ((gx >= get_image_width(dst)) | (gy >= get_image_height(dst))) 
      return; 

__local float4 la[wgs+2][wgs+2]; 

la[lx+1][ly+1] = read_imagef(src,smp,(int2)(gx,gy));  

if(lx == 0){ 
    la[lx][ly+1] = read_imagef(src,smp,(int2)(gx-1,gy)); 
    if(ly == 0) 
     la[lx+1][ly] = read_imagef(src,smp,(int2)(gx-1,gy)); 
    if(ly == wgs) 
     la[lx+1][ly+2] = read_imagef(src,smp,(int2)(gx+1,gy)); 
} 
else if(lx == wgs){ 
    la[lx+2][ly+1] = read_imagef(src,smp,(int2)(gx+1,gy)); 
    if(ly == 0) 
     la[lx+1][ly] = read_imagef(src,smp,(int2)(gx-1,gy)); 
    if(ly == wgs) 
     la[lx+1][ly+2] = read_imagef(src,smp,(int2)(gx+1,gy)); 
} 
else if(ly == 0) 
    la[lx+1][ly] = read_imagef(src,smp,(int2)(gx-1,gy)); 
else if(ly == wgs) 
    la[lx+1][ly+2] = read_imagef(src,smp,(int2)(gx+1,gy)); 


barrier(CLK_LOCAL_MEM_FENCE); //----------------------- mem barrier 

    float4 s0 = la[ lx-1][ly-1 ]; 
    float4 s1 = la[ lx ][ly-1 ]; 
    float4 s2 = la[ lx+1][ly-1 ]; 
    float4 s3 = la[ lx-1][ly ]; 
    float4 s4 = la[ lx ][ly ]; 
    float4 s5 = la[ lx+1][ly ]; 
    float4 s6 = la[ lx-1][ly+1 ]; 
    float4 s7 = la[ lx ][ly+1 ]; 
    float4 s8 = la[ lx+1][ly+1 ]; 

启动分拣 这方面的工作得很好不用担心它

// stage0 
    cas3(s1, s2); 
    cas3(s4, s5); 
    cas3(s7, s8); 

    // 1 
    cas3(s0, s1); 
    cas3(s3, s4); 
    cas3(s6, s7); 

    // 2 
    cas3(s1, s2); 
    cas3(s4, s5); 
    cas3(s7, s8); 

    // 3/4 
    cas3(s3, s6); 
    cas3(s4, s7); 
    cas3(s5, s8); 
    cas3(s0, s3); 

    cas3(s1, s4); 
    cas3(s2, s5); 
    cas3(s3, s6); 

    cas3(s4, s7); 
    cas3(s1, s3); 

    cas3(s2, s6); 
    cas3(s2, s3); 
    cas3(s4, s6); 

    cas3(s3, s4); 

结束分拣

write_imagef(dst, (int2) (gx, gy), s4); 
} 

回答

0

一对夫妇的建议,随机试验:

  • 如果以后出现障碍,应尽早使用提前退货:全部工作组中的工作项应该执行屏障,并且如果某些工作项已返回,则内核将挂起。

  • 纹理缓存在这种情况下通常至少与本地内存一样快。尝试用简单的read_imagef调用替换s0,...,s8初始化。然后尝试不同的工作组维度。

您在运行什么硬件?

+0

我在i7和660m上运行它 –