关于可能定义以下内容的任何想法？
ahcarpenter@ahcarpenter-laptop:~$ g++ objectmarker.o -o objectmarker
objectmarker.o: In function `on_mouse(int, int, int, int, void*)':
objectmarker.cpp:(.text+0x12f): undefined reference to `cvCloneImage'
objectmarker.cpp:(.text+0x1d1): undefined reference to `cvRectangle'
objectmarker.cpp:(.text+0x1ea): undefined reference to `cvShowImage'
objectmarker.cpp:(.text+0x1f4): undefined reference to `cvReleaseImage'
objectmarker.o: In function `main':
objectmarker.cpp:(.text+0x391): undefined reference to `cvNamedWindow'
objectmarker.cpp:(.text+0x3aa): undefined reference to `cvSetMouseCallback'
objectmarker.cpp:(.text+0x4da): undefined reference to `cvLoadImage'
objectmarker.cpp:(.text+0x50f): undefined reference to `cvShowImage'
objectmarker.cpp:(.text+0x519): undefined reference to `cvWaitKey'
objectmarker.cpp:(.text+0x53f): undefined reference to `cvReleaseImage'
objectmarker.cpp:(.text+0x54e): undefined reference to `cvDestroyWindow'
objectmarker.cpp:(.text+0xd7f): undefined reference to `cvReleaseImage' …我的代码就像这些一样简单:
#include <iostream>
using namespace std;
//Some codes here...
bool somefunction(){
cout<<"???";
return false;
}
这就是我得到的:
error C2143: syntax error: missing ';' before 'return'; error C2001: newline is constant;
而且,如果我"???"换成像"细胞"这样的英文版,那就完美了;
我有一个bmp格式的图像,大小为512*512.我想计算值大于11的像素数,然后找到这些像素的平均值.这是我的代码.我不知道问题是什么,但像素值的总和是错误的,它总是255.我尝试了不同的图像.
你能帮我解决一下吗？
A=imread('....bmp');
sum=0; count=0;
for i=1:512
for j=1:512
if (A(i,j)>=11)
sum=sum+A(i,j);
count=count+1;
end
end
end
disp('Number of pixels grater than or equal to 11')
disp(count)
disp('sum')
disp(sum)
disp('Average')
Avrg=sum/count;
disp(Avrg)
我有一个数据表,如下所示.请注意,keyID可以是重复的.我已经在矢量结构中收集了以下数据,并对其进行了排序.
struct myData {
int keyID;
int value;
}
vector<myData> vecReadFromFile;
现在用户输入一个特定的keyID,我必须检查该值是否在向量中退出,如果退出,我必须返回该值.如果不是,我必须检查它落在哪个值之间,例如,如果用户输入120030,值在120028和120039之间,我应该得到这些值的索引,即此示例中的lowerIndex和upperIndex为'2'和'3'(作为向量index从0开始)
如果用户输入较少的keyID,即120001则不返回任何值.类似地,用户输入的keyID大于最后一个键值,然后返回不同的错误代码.
基本上我想有效地找到给定键值的索引范围.我添加的代码似乎不适用于上面的例子我提到什么是bug？
我可以改变逻辑以使用STL提供的算法.请建议.
我们如何在C++中有效地实现这种算法？请求示例代码作为函数.请注意,我将在我的项目中多次调用函数,因此它必须有效.
keyID Value
120002 10
120025 20
120028 25
120039 30
120042 -
120048 40
120052 50
120112 60
120117 70
120123 70
120126 80
120130 90
我这里有一些代码
//==========================================================================
// FindBounds
bool FindBounds(const KEY& cTarget, UINT& uLower, UINT& uUpper)
{
uLower = -1;
uUpper = -1;
// start with full range of data.
uLower = 0;
uUpper = m_uCount-1; // Here I have m_uCount as …class loss(Function):
@staticmethod
def forward(ctx,x,INPUT):
batch_size = x.shape[0]
X = x.detach().numpy()
input = INPUT.detach().numpy()
Loss = 0
for i in range(batch_size):
t_R_r = input[i,0:4]
R_r = t_R_r[np.newaxis,:]
t_R_i = input[i,4:8]
R_i = t_R_i[np.newaxis,:]
t_H_r = input[i,8:12]
H_r = t_H_r[np.newaxis,:]
t_H_i = input[i,12:16]
H_i = t_H_i[np.newaxis,:]
t_T_r = input[i, 16:32]
T_r = t_T_r.reshape(4,4)
t_T_i = input[i, 32:48]
T_i = t_T_i.reshape(4,4)
R = np.concatenate((R_r, R_i), axis=1)
H = np.concatenate((H_r, H_i), axis=1)
temp_t1 = np.concatenate((T_r,T_i),axis=1)
temp_t2 = np.concatenate((-T_i,T_r),axis=1)
T = np.concatenate((temp_t1,temp_t2),axis=0)
phi_r = …