我想模拟CPU上CUDA双线性插值的行为,但我发现tex2D
的返回值似乎不适合bilinear formula。C/C++和CUDA中的双线性插值
我想,从float
到9
-bit固定点格式与8
位小数值铸造插值系数会导致不同的值。
据转换fomula [2, line 106],所述转换的结果将是相同的作为输入float
当coeffient是1/2^n
,与n=0,1,..., 8
,但我仍然(不总是)接收怪异值。
下面我报告一个怪异值的例子。在这种情况下,奇怪的值总是发生在id = 2*n+1
,有谁能告诉我为什么?
Src的阵列:
Src[0][0] = 38;
Src[1][0] = 39;
Src[0][1] = 118;
Src[1][1] = 13;
纹理定义:
static texture<float4, 2, cudaReadModeElementType> texElnt;
texElnt.addressMode[0] = cudaAddressModeClamp;
texElnt.addressMode[1] = cudaAddressModeClamp;
texElnt.filterMode = cudaFilterModeLinear;
texElnt.normalized = false;
核函数:
static __global__ void kernel_texElnt(float* pdata, int w, int h, int c, float stride/*0.03125f*/) {
const int gx = blockIdx.x*blockDim.x + threadIdx.x;
const int gy = blockIdx.y*blockDim.y + threadIdx.y;
const int gw = gridDim.x * blockDim.x;
const int gid = gy*gw + gx;
if (gx >= w || gy >= h) {
return;
}
float2 pnt;
pnt.x = (gx)*(stride)/*1/32*/;
pnt.y = 0.0625f/*1/16*/;
float4 result = tex2D(texElnt, pnt.x + 0.5, pnt.y + 0.5f);
pdata[gid*3 + 0] = pnt.x;
pdata[gid*3 + 1] = pnt.y;
pdata[gid*3 + 2] = result.x;
}
CUDA的双线性结果
id pnt.x pnt.y tex2D
0 0.00000 0.0625 43.0000000
1 0.03125 0.0625 42.6171875
2 0.06250 0.0625 42.6484375
3 0.09375 0.0625 42.2656250
4 0.12500 0.0625 42.2968750
5 0.15625 0.0625 41.9140625
6 0.18750 0.0625 41.9453125
7 0.21875 0.0625 41.5625000
8 0.25000 0.0625 41.5937500
9 0.28125 0.0625 41.2109375
0 0.31250 0.0625 41.2421875
10 0.34375 0.0625 40.8593750
11 0.37500 0.0625 40.8906250
12 0.40625 0.0625 40.5078125
13 0.43750 0.0625 40.5390625
14 0.46875 0.0625 40.1562500
15 0.50000 0.0625 40.1875000
16 0.53125 0.0625 39.8046875
17 0.56250 0.0625 39.8359375
18 0.59375 0.0625 39.4531250
19 0.62500 0.0625 39.4843750
20 0.65625 0.0625 39.1015625
21 0.68750 0.0625 39.1328125
22 0.71875 0.0625 38.7500000
23 0.75000 0.0625 38.7812500
24 0.78125 0.0625 38.3984375
25 0.81250 0.0625 38.4296875
26 0.84375 0.0625 38.0468750
27 0.87500 0.0625 38.0781250
28 0.90625 0.0625 37.6953125
29 0.93750 0.0625 37.7265625
30 0.96875 0.0625 37.3437500
31 1.00000 0.0625 37.3750000
CPU结果:
// convert coefficient ((1-α)*(1-β)), (α*(1-β)), ((1-α)*β), (α*β) to fixed point format
id pnt.x pnt.y tex2D
0 0.00000 0.0625 43.00000000
1 0.03125 0.0625 43.23046875
2 0.06250 0.0625 42.64843750
3 0.09375 0.0625 42.87890625
4 0.12500 0.0625 42.29687500
5 0.15625 0.0625 42.52734375
6 0.18750 0.0625 41.94531250
7 0.21875 0.0625 42.17578125
8 0.25000 0.0625 41.59375000
9 0.28125 0.0625 41.82421875
0 0.31250 0.0625 41.24218750
10 0.34375 0.0625 41.47265625
11 0.37500 0.0625 40.89062500
12 0.40625 0.0625 41.12109375
13 0.43750 0.0625 40.53906250
14 0.46875 0.0625 40.76953125
15 0.50000 0.0625 40.18750000
16 0.53125 0.0625 40.41796875
17 0.56250 0.0625 39.83593750
18 0.59375 0.0625 40.06640625
19 0.62500 0.0625 39.48437500
20 0.65625 0.0625 39.71484375
21 0.68750 0.0625 39.13281250
22 0.71875 0.0625 39.36328125
23 0.75000 0.0625 38.78125000
24 0.78125 0.0625 39.01171875
25 0.81250 0.0625 38.42968750
26 0.84375 0.0625 38.66015625
27 0.87500 0.0625 38.07812500
28 0.90625 0.0625 38.30859375
29 0.93750 0.0625 37.72656250
30 0.96875 0.0625 37.95703125
31 1.00000 0.0625 37.37500000
我留下一个简单的代码上my github [3],运行你会在D:\
有两个文件程序后。
编辑2014年1月20日
我运行不同的增量项目,发现tex2D
规范“时alpha
乘以beta
小于0.00390625
的tex2D
回报不匹配,双线性插值公式“
您可以添加其他人可以编译和运行的最短完整示例吗? – talonmies
感谢您的建议@talonmies,我提供了示例代码的链接。 – user1995868