调用静态内联函数

Calling a static inline function

本文关键字：函数 静态 调用      更新时间：2023-10-16

如何调用此static inline函数？

static inline int xGradient(uchar* image, int x, int y)
{
return ((int)(image[y-1, x-1])) +
2*image[y, x-1] +
image[y+1, x-1] -
image[y-1, x+1] -
2*image[y, x+1] -
image[y+1, x+1];
}
static inline int yGradient(uchar* image, int x, int y)
{
return ((int)(image[y-1, x-1])) +
2*image[y-1, x] +
image[y-1, x+1] -
image[y+1, x-1] -
2*image[y+1, x] -
image[y+1, x+1];

调用函数时遇到麻烦。我这样称呼它：

gx = xGradient(&data[ii], x,y);
gy = yGradient(&data[ii], x,y);
sum = abs(gx) + abs(gy);
sum = sum > 255 ? 255:sum;
sum = sum < 0 ? 0 : sum;

我没有得到gx和gy的结果。请帮我在上面的程序中计算gx和gy。

这是我的代码

#include<iostream>
#include<omp.h>
#include<ctime>
#include<cmath>
#include<opencv2/imgproc/imgproc.hpp>
#include<opencv2/highgui/highgui.hpp>
using namespace std;
using namespace cv;
static inline int xGradient(uchar* image, int x, int y)
{
return ((int)(image[y-1, x-1])) +
2*image[y, x-1] +
image[y+1, x-1] -
image[y-1, x+1] -
2*image[y, x+1] -
image[y+1, x+1];
}
static inline int yGradient(uchar* image, int x, int y)
{
return ((int)(image[y-1, x-1])) +
2*image[y-1, x] +
image[y-1, x+1] -
image[y+1, x-1] -
2*image[y+1, x] -
image[y+1, x+1];
}
int main()  
{
Mat src, grey, grey2, dst;
clock_t start, finish;
int gx, gy, sum;
size_t total;
int sizes[2];
start = clock();
src= imread("E:/sobel/Debug/view_sea.bmp");
imwrite("E:/sobel/Debug/Serial/Citra Asli.bmp", src );
cvtColor(src,grey,CV_BGR2GRAY);
imwrite("E:/sobel/Debug/Serial/Grayscale.bmp", grey );
dst = grey.clone();
if( !grey.data )
{
return -1;
}
total=grey.total();
cv::Size s = grey.size();
sizes[0] = s.height;
sizes[1] = s.width;
cout << "citra terdiri dari " << total << " piksel dengan ukuran " << sizes[0] << " x " << sizes[1] << " piksel" << endl;
int starty=(grey.rows);
if(starty==0)
{starty=1;}
int stopy=(grey.rows);
if(stopy>grey.rows - 1)
{stopy=grey.rows - 1;}
int ii=grey.cols;
uchar* data=grey.data;
for(int y = starty; y < stopy; y++)
{
ii++;
for(int x = 1; x < sizes[1] - 1; x++)
{
gx = xGradient(&data[ii], x,y);
gy = yGradient(&data[ii], x,y);
sum = abs(gx) + abs(gy);
sum = sum > 255 ? 255:sum;
sum = sum < 0 ? 0 : sum;
data[ii] = sum;
ii++;
}
ii++;
}
finish = clock();
imwrite( "E:/sobel/Debug/Serial/Output sobel dengan Serial.bmp", src);
cout << "Waktu Eksekusi Deteksi Tepi Serial adalah : " << float(finish-  start)/CLOCKS_PER_SEC << " detik" << endl;
return 0;
}

我在这个代码中出错了

int ii=grey.cols;
uchar* data=grey.data;
for(int y = starty; y < stopy; y++)
{
ii++;
for(int x = 1; x < sizes[1] - 1; x++)
{
gx = xGradient(&data[ii], x,y);
gy = yGradient(&data[ii], x,y);
sum = abs(gx) + abs(gy);
sum = sum > 255 ? 255:sum;
sum = sum < 0 ? 0 : sum;
data[ii] = sum;
ii++;
}
ii++;
}

我认为，您混淆了python/numpy和c++语法。

而image[y-1, x-1]将在python中完成正确的工作(给定一个2d numpy数组)，

在c++中，您只得到一个1d的uchar数组，它可以归结为image[x-1]。可能不是你所期望的。

要做到这一点，您的函数还需要一个额外的参数，1行的大小(宽度)：

static inline int xGradient(uchar* image, int x, int y, int W)
{
return (
image[ W*(y-1) + (x-1)] +
2*image[ W*(y) + (x-1)] +
image[ W*(y+1) + (x-1)] -
image[ W*(y-1) + (x+1)] -
2*image[ W*(y) + (x+1)] -
image[ W*(y+1) + (x+1)] );
}

但是，既然我们在opencv中，为什么不使用Mat对象本身，而不是原始字节：

static inline int xGradient(const Mat & img, int x, int y)
{
return (
img.at<uchar>( (y-1) , (x-1) ) +
2*img.at<uchar>( (y) , (x-1) ) +
img.at<uchar>( (y+1) , (x-1) ) -
img.at<uchar>( (y-1) , (x+1) ) -
2*img.at<uchar>( (y) , (x+1) ) -
img.at<uchar>( (y+1) , (x+1) ) );
}
// and call it : 
Mat img;
int xg = xGradient(img,x,y);

记住，当应用这个时，你必须在图像中留出一个1像素的边界，否则你就越界了。。。

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