OpenCV SURF 比较描述符
以下代码片段来自 OpenCV find_obj.cpp，它是使用 SURF 的演示，
double
compareSURFDescriptors( const float* d1, const float* d2, double best, int length )
{
double total_cost = 0;
assert( length % 4 == 0 );
int i;
for( i = 0; i best )
break;
}
return total_cost;
}
据我所知它检查欧几里得距离，我不明白的是为什么会这样4人一组做吗？为什么不一次性计算出全部内容呢？
Folowing snippet is from OpenCV find_obj.cpp which is demo for using SURF,
double
compareSURFDescriptors( const float* d1, const float* d2, double best, int length )
{
double total_cost = 0;
assert( length % 4 == 0 );
int i;
for( i = 0; i best )
break;
}
return total_cost;
}
As far as I can tell it checking the euclidian distance, what I do not understand is why is it doing it in groups of 4? Why not calculate the whole thing at once?
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通常这样做是为了使 SSE 优化成为可能。 SSE 寄存器有 128 位长，可以包含 4 个浮点数，因此您可以使用一条指令并行执行 4 次减法。
另一个好处是：只有在每第四个差异之后才需要检查循环计数器。即使编译器不利用生成 SSE 代码的机会，这也会使代码更快。例如，VS2008没有，即使使用-O2也没有：
double t0 = d1[i] - d2[i]; 00D91666 fld dword ptr [edx-0Ch] 00D91669 fsub dword ptr [ecx-4] double t1 = d1[i+1] - d2[i+1]; 00D9166C fld dword ptr [ebx+ecx] 00D9166F fsub dword ptr [ecx] double t2 = d1[i+2] - d2[i+2]; 00D91671 fld dword ptr [edx-4] 00D91674 fsub dword ptr [ecx+4] double t3 = d1[i+3] - d2[i+3]; 00D91677 fld dword ptr [edx] 00D91679 fsub dword ptr [ecx+8] total_cost += t0*t0 + t1*t1 + t2*t2 + t3*t3; 00D9167C fld st(2) 00D9167E fmulp st(3),st 00D91680 fld st(3) 00D91682 fmulp st(4),st 00D91684 fxch st(2) 00D91686 faddp st(3),st 00D91688 fmul st(0),st 00D9168A faddp st(2),st 00D9168C fmul st(0),st 00D9168E faddp st(1),st 00D91690 faddp st(2),stUsually things like this are done for making SSE optimizations possible. SSE registers are 128 bits long and can contain 4 floats, so you can do the 4 subtractions using one instruction, parallelly.
Another upside: you have to check the loop counter only after every fourth difference. That makes the code faster even if the compiler doesn't use the opportunity to generate SSE code. For example, VS2008 didn't, not even with -O2:
double t0 = d1[i] - d2[i]; 00D91666 fld dword ptr [edx-0Ch] 00D91669 fsub dword ptr [ecx-4] double t1 = d1[i+1] - d2[i+1]; 00D9166C fld dword ptr [ebx+ecx] 00D9166F fsub dword ptr [ecx] double t2 = d1[i+2] - d2[i+2]; 00D91671 fld dword ptr [edx-4] 00D91674 fsub dword ptr [ecx+4] double t3 = d1[i+3] - d2[i+3]; 00D91677 fld dword ptr [edx] 00D91679 fsub dword ptr [ecx+8] total_cost += t0*t0 + t1*t1 + t2*t2 + t3*t3; 00D9167C fld st(2) 00D9167E fmulp st(3),st 00D91680 fld st(3) 00D91682 fmulp st(4),st 00D91684 fxch st(2) 00D91686 faddp st(3),st 00D91688 fmul st(0),st 00D9168A faddp st(2),st 00D9168C fmul st(0),st 00D9168E faddp st(1),st 00D91690 faddp st(2),st我认为这是因为每个分区都有 4 个数字。总共 4x4x4 子区域，形成 64 长度向量。所以它基本上得到了两个子区域之间的差异。
I think it is because for each subregion we get 4 numbers. Totally 4x4x4 subregions making 64 length vector. So its basically getting the difference between 2 sub regions.