目录
2、输入4*4*1,卷积核3*3*1,步长1*1,输出2*2*1
3、输入4*4*1,卷积核2*2*1,步长2*2,输出2*2*1
4、输入4*4*1,卷积核3*3*1,步长1*1,padding为1*1,输出2*2*1
5、输入4*4*2,卷积核3*3*2,步长1*1,padding为1*1,输出2*2*1
6、输入4*4*1,卷积核3*3*1,卷积核个数2,步长1*1,padding为1*1,输出2*2*1
7、输入4*4*2,卷积核3*3*2,步长1*1,padding为1*1,膨胀系数2,输出2*2*2
参考资料:
(图解)一步一步使用CPP实现深度学习中的卷积 - GiantPandaCV
其实在上述链接里已经写的非常详细了,为了便于后续理解和学习,所以写个理解的备忘录。
一、参数说明和计算公式
1、符号约定
cpp
F[] 为输入;
width 为输入的宽;
height 为输入的高;
channel 为输入的通道;
K[] 为 kernel;
kSizeX 为 kernel 的宽;
kSizeY 为 kernel 的高;
filters 为 kernel 的个数;
padX 为水平方向的填充;
padY 为垂直方向的填充;
strideX 为水平方向的步长;
strideY 为垂直方向的步长;
O[] 为输出;
outW 为输出的宽;
outH 为输出的高;
outChannel 为输出的通道;
2、输出大小计算公式
cpp
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
公式说明:
-
height + 2 * padY
: 首先,将输入图像的高度加上两倍的填充量,这是因为填充是对称地加在图像的顶部和底部。 -
dilationY * (kSizeY - 1)
: 这是计算卷积核中除了中心元素外,其他元素之间的总间隔。kSizeY - 1
表示卷积核的高度减去中心元素,乘以膨胀系数dilationY
,得到这些元素之间的间隔数。 -
dilationY * (kSizeY - 1) + 1
: 将上述间隔数加上1,是因为我们还要包括卷积核的中心元素。 -
(height + 2 * padY - (dilationY * (kSizeY - 1) + 1))
: 这个表达式计算了在考虑填充和膨胀后,输入图像实际上被卷积核覆盖的区域的高度。 -
/ strideY
: 将上述计算得到的高度除以步长,得到在垂直方向上卷积核可以覆盖的步数。 -
+ 1
: 最后,由于输出的高度至少要有一个元素,即使计算结果为0,我们也需要加1,以确保输出尺寸至少为1。
二、不同类型的卷积
1、输入3*3*1,卷积核3*3*1,输出1*1*1
(1)实现代码
封装成demo0(),main()函数可直接调用
cpp
void demo0()
{
float F[] = {1,2,3,4,5,6,7,8,9};
float K[] = {1,2,3,4,5,6,7,8,9};
float O = 0;
int width = 3;
int height = 3;
int kSizeX = 3;
int kSizeY = 3;
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
O+=K[m*kSizeX+n]*F[m*width+n];
}
}
std::cout<<O<<" ";
}
(2)代码说明
- 前面都是定义输入矩阵和卷积核,同时定义一个浮点数变量O并初始化为0,用于累加结果。
- 两个嵌套循环遍历卷积核的每个元素:
- for(int m=0; m<kSizeY; m++):外层循环遍历卷积核的高度。
- for(int n=0; n<kSizeX; n++):内层循环遍历卷积核的宽度。
- 嵌套循环内部,O += K[m*kSizeX + n] * F[m*width + n];用来执行卷积操作。这行代码将卷积核的当前元素 K[m*kSizeX + n] 与输入矩阵的对应元素 F[m*width + n] 相乘,并将结果累加到 O 中【其中位置是相同的,m只是用来决定以kSize为单位的轮次】。
2、输入4*4*1,卷积核3*3*1,步长1*1,输出2*2*1
(1)实现代码
cpp
void demo1()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 0;
int padY = 0;
int dilationX = 1;
int dilationY = 1;
int strideX = 1;
int strideY = 1;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
O[i*outW+j]+=K[m*kSizeX+n]*F[(m+i)*width+(n+j)];
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
(2)代码说明
- 前面都是定义输入矩阵和卷积核,同时定义一个浮点数变量O并初始化【和前面后面一样,都是先计算出来输出结果的大小,然后再执行】,用于累加结果。
- 四个嵌套循环开始进行卷积计算:
- 外层两个循环遍历输出特征图的每个位置。
- 内层两个循环遍历卷积核的每个元素。
- 在内层循环中,计算卷积核覆盖的输入特征图区域的加权和,并将结果累加到输出特征图的相应位置。
3、输入4*4*1,卷积核2*2*1,步长2*2,输出2*2*1
(1)实现代码
cpp
void demo2()
{
// (height + 2 * paddingY - (dilationY * (kSizeY - 1) + 1)) / strideY + 1;
// (width + 2 * paddingX - (dilationX * (kSizeX - 1) + 1)) / strideX + 1;
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4};
//float K[] = {1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 0;
int padY = 0;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 2;
int kSizeY = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
O[i*outW+j]+=K[m*kSizeX+n]*F[(m+i*strideY)*width+(n+j*strideX)];
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
这个和上一段代码解释相同,只不过步长和卷积核大小有改变。
4、输入4*4*1,卷积核3*3*1,步长1*1,padding为1*1,输出2*2*1
(1)实现代码
cpp
void demo3()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
//考虑边界强情况
if((n+j*strideX-padX)>-1&&(m+i*strideY-padY>-1)&&(n+j*strideX-padX)<=width&&(m+i*strideY-padY>-1)<=height)
{
fVal = F[(m+i*strideY-padX)*width+(n+j*strideX-padY)];
}
O[i*outW+j]+=K[m*kSizeX+n]*fVal;
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
(2)代码说明
按照正常计算输出规模公式计算的话,应该输出是4*4*1。但是考虑到边界墙问题。
if
语句块用来检查当前卷积核的 m
和 n
索引是否在输入特征图的边界内。这里考虑了填充(padding)的情况:
(n+j*strideX - padX)
计算了卷积核的左边界在输入特征图上的索引。(m+i*strideY - padY)
计算了卷积核的上边界在输入特征图上的索引。- 条件
> -1
确保索引不会是负数。 - 条件
<= width
和<= height
确保索引不会超出输入特征图的边界。
5、输入4*4*2,卷积核3*3*2,步长1*1,padding为1*1,输出2*2*1
其中依旧考虑了边界墙问题。
(1)实现代码
cpp
void demo4()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int channel = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for (int c = 0; c < channel; ++c)
{
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
if((n+j*strideX-padX)>-1&&(m+i*strideY-padY>-1)&&(n+j*strideX-padX)<=width&&(m+i*strideY-padY>-1)<=height)
{
fVal = F[c*width*height + (m+i*strideY-padX)*width+(n+j*strideX-padY)];
}
O[i*outW+j]+=K[c*kSizeX*kSizeY+m*kSizeX+n]*fVal;
}
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
6、输入4*4*1,卷积核3*3*1,卷积核个数2,步长1*1,padding为1*1,输出2*2*1
(1)实现代码
cpp
void demo5()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,
1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9
};
float O[] = {0,0,0,0,0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int channel = 2;
int filters = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
int outC = filters;
for (int oc = 0; oc < outC; ++oc)
{
for (int c = 0; c < channel; ++c)
{
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
if((n+j*strideX-padX)>-1&&(m+i*strideY-padY>-1)&&(n+j*strideX-padX)<=width&&(m+i*strideY-padY>-1)<=height)
{
fVal = F[c*width*height + (m+i*strideY-padX)*width+(n+j*strideX-padY)];
}
O[oc*outH*outW+i*outW+j]+=K[oc*outC*kSizeX*kSizeY+c*kSizeX*kSizeY+m*kSizeX+n]*fVal;
}
}
}
}
}
}
for (int oc = 0; oc < outC; ++oc)
{
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[oc*outH*outW+i*outW+j]<<" ";
}
std::cout<<std::endl;
}
std::cout<<std::endl<<std::endl;
}
}
(2)代码说明
-
六个嵌套循环实现了卷积操作:
- 最外层循环遍历输出特征图的每个通道。
- 第二个循环遍历输入特征图的每个通道。
- 第三个和第四个循环遍历输出特征图的每个元素。
- 最内层两个循环遍历每个卷积核的元素。
-
在最内层循环中,首先检查当前卷积核的索引是否在输入特征图的边界内,如果是,则从输入特征图中取出相应的元素
fVal
。 -
然后,将
fVal
与对应卷积核的元素相乘,并将结果累加到输出特征图的对应位置。
7、输入4*4*2,卷积核3*3*2,步长1*1,padding为1*1,膨胀系数2,输出2*2*2
cpp
void demo6()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,
1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9
};
float O[] = {0,0,0,0,0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 2;
int dilationY = 2;
int strideX = 1;
int strideY = 1;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int channel = 2;
int filters = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
int outC = filters;
for (int oc = 0; oc < outC; ++oc)
{
for (int c = 0; c < channel; ++c)
{
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
if( ((n+j*strideX)*dilationX-padX)>-1 && ((m+i*strideY)*dilationY-padY)>-1&&
((n+j*strideX)*dilationX-padX)<=width && ((m+i*strideY)*dilationY-padY>-1)<=height)
{
fVal = F[c*width*height + ((m+i*strideY)*dilationX-padX)*width+((n+j*strideX)*dilationY-padY)];
}
O[oc*outH*outW+i*outW+j]+=K[oc*outC*kSizeX*kSizeY+c*kSizeX*kSizeY+m*kSizeX+n]*fVal;
}
}
}
}
}
}
for (int oc = 0; oc < outC; ++oc)
{
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[oc*outH*outW+i*outW+j]<<" ";
}
std::cout<<std::endl;
}
std::cout<<std::endl;
}
}
三、可执行的卷积操作代码
cpp
#include <iostream>
void demo0() //输入3*3*1,卷积核3*3*1,输出为1*1*1
{
float F[] = {1,2,3,4,5,6,7,8,9};
float K[] = {1,2,3,4,5,6,7,8,9};
float O = 0;
int width = 3;
int height = 3;
int kSizeX = 3;
int kSizeY = 3;
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
O+=K[m*kSizeX+n]*F[m*width+n];
}
}
std::cout<<O<<" ";
}
void demo1() //输入4*4*1,卷积核3*3*1,步长为1*1,no padding,输出为1*1*1
{
//计算公式
// (height + 2 * paddingY - (dilationY * (kSizeY - 1) + 1)) / strideY + 1;
// (width + 2 * paddingX - (dilationX * (kSizeX - 1) + 1)) / strideX + 1;
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 0;
int padY = 0;
int dilationX = 1;
int dilationY = 1;
int strideX = 1;
int strideY = 1;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
O[i*outW+j]+=K[m*kSizeX+n]*F[(m+i)*width+(n+j)];
/* int inputIndex = (i * strideY + m) * width + (j * strideX + n);
O[i * outW + j] += K[m * kSizeX + n] * F[inputIndex]; */
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
void demo2()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4};
//float K[] = {1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 0;
int padY = 0;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 2;
int kSizeY = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
O[i*outW+j]+=K[m*kSizeX+n]*F[(m+i*strideY)*width+(n+j*strideX)];
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
void demo3()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
//float K[] = {1,2,3,4};
float K[] = {1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
//考虑边界强情况
if((n+j*strideX-padX)>-1&&(m+i*strideY-padY>-1)&&(n+j*strideX-padX)<=width&&(m+i*strideY-padY>-1)<=height)
{
fVal = F[(m+i*strideY-padX)*width+(n+j*strideX-padY)];
}
O[i*outW+j]+=K[m*kSizeX+n]*fVal;
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
void demo4()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
//float K[] = {1,2,3,4};
float K[] = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int channel = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for (int c = 0; c < channel; ++c)
{
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
if((n+j*strideX-padX)>-1&&(m+i*strideY-padY>-1)&&(n+j*strideX-padX)<=width&&(m+i*strideY-padY>-1)<=height)
{
fVal = F[c*width*height + (m+i*strideY-padX)*width+(n+j*strideX-padY)];
}
O[i*outW+j]+=K[c*kSizeX*kSizeY+m*kSizeX+n]*fVal;
}
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
void demo5()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
//float K[] = {1,2,3,4};
float K[] = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,
1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9
};
float O[] = {0,0,0,0,0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int channel = 2;
int filters = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
int outC = filters;
for (int oc = 0; oc < outC; ++oc)
{
for (int c = 0; c < channel; ++c)
{
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
if((n+j*strideX-padX)>-1&&(m+i*strideY-padY>-1)&&(n+j*strideX-padX)<=width&&(m+i*strideY-padY>-1)<=height)
{
fVal = F[c*width*height + (m+i*strideY-padX)*width+(n+j*strideX-padY)];
}
O[oc*outH*outW+i*outW+j]+=K[oc*channel*kSizeX*kSizeY+c*kSizeX*kSizeY+m*kSizeX+n]*fVal;
}
}
}
}
}
}
for (int oc = 0; oc < outC; ++oc)
{
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[oc*outH*outW+i*outW+j]<<" ";
}
std::cout<<std::endl;
}
std::cout<<std::endl<<std::endl;
}
}
void demo6()
{
// (height + 2 * paddingY - (dilationY * (kSizeY - 1) + 1)) / strideY + 1;
// (width + 2 * paddingX - (dilationX * (kSizeX - 1) + 1)) / strideX + 1;
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
//float K[] = {1,2,3,4};
float K[] = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,
1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9
};
float O[] = {0,0,0,0,0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 2;
int dilationY = 2;
int strideX = 1;
int strideY = 1;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int channel = 2;
int filters = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
int outC = filters;
for (int oc = 0; oc < outC; ++oc)
{
for (int c = 0; c < channel; ++c)
{
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
if( ((n+j*strideX)*dilationX-padX)>-1 && ((m+i*strideY)*dilationY-padY)>-1&&
((n+j*strideX)*dilationX-padX)<=width && ((m+i*strideY)*dilationY-padY>-1)<=height)
{
fVal = F[c*width*height + ((m+i*strideY)*dilationX-padX)*width+((n+j*strideX)*dilationY-padY)];
}
O[oc*outH*outW+i*outW+j]+=K[oc*channel*kSizeX*kSizeY+c*kSizeX*kSizeY+m*kSizeX+n]*fVal;
}
}
}
}
}
}
for (int oc = 0; oc < outC; ++oc)
{
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[oc*outH*outW+i*outW+j]<<" ";
}
std::cout<<std::endl;
}
std::cout<<std::endl;
}
}
int main(int argc, char *argv[])
{
//demo0();
demo1();
//demo2();
//demo3();
//demo4();
//demo5();
//demo6();
}
(1)将上述文件为my_code.cc源文件。
如果要使用不同类型的demo,可以直接在main函数中调用。
(2)然后在终端运行:
bash
g++ -o my_code my_code.cc
此时已编译出可执行的二进制my_code。
(3)执行my_code文件
bash
./my_code
最后说明:其实我个人觉得这种方式使用起来除了简单易懂,但在实际操作中并不可取。主要是所有的变量值都需要在demo()函数中手工赋值,耦合性和实用性较差。但是目前也没有更好的实测,等过阵子我的程序写完后,再和大家分享。