将opencv读取图像传递给dll c ++函数

Passing opencv read image to dll c++ function

我在从 C++ 函数读取像素信息时遇到问题,我可以读取所有其他参数,因此我确信我的连接正常。 我的python代码如下

real_img = cv2.imread("215107.jpg", 0)
n=real_image.ctypes.data_as(ctypes.POINTER(ctypes.c_uint8))
hllDll=WinDLL(name)
hllDll.transform(n,ctypes.c_int(myimage.shape[0]), ctypes.c_int(myimage.shape[1]), ctypes.c_int(angle))

我也试过了

src1 = real_image.ctypes.data_as(ctypes.c_char_p)
hllDll.transform(src1,ctypes.c_int(myimage.shape[0]), ctypes.c_int(myimage.shape[1]), ctypes.c_int(angle))

src2=ctypes.c_void_p(real_img.ctypes.data)
hllDll.transform(src2,ctypes.c_int(myimage.shape[0]), ctypes.c_int(myimage.shape[1]), ctypes.c_int(angle))

但失败了 我的 C++ 函数如下:

unsigned int transform(int** a, int row,int col, int ang){
for (int i = 0; i < 4; i++) {
        cout << a[i] << endl;
    }
...}

当我尝试读取 a 时,它们都给我相同的大数字,例如

1212121212121212
1212121212121212
1413131211111010
1012141311101113

虽然如果我通过 python 访问,前 5 个数字应该是 18,但我很困惑是什么导致了这种情况

编辑: 我也试过 cout << a[i][1] << endl; 但是 OSError: exception: access violation reading 0xFFFFFFFFFFFFFFFF 返回

您正在打印每一行的指针。如果你想打印第一行的元素,你应该这样做:

unsigned int transform(int** a, int row,int col, int ang){
    for (int i = 0; i < 4; i++) {
        cout << a[0][i] << endl; // add [0] to access the first row
    }
    ...
}

您传递的是 np.uint8 类型的数组,等效的 C 指针是 unsigned char*.

int** a替换为unsigned char* a


为了测试,您可以填写 real_image 的前 4 个元素:

real_image[0, 0] = 1
real_image[0, 1] = 2
real_image[0, 2] = 3
real_image[0, 3] = 4

hllDll.transform(n,ctypes.c_int(real_image.shape[0]), ctypes.c_int(real_image.shape[1]), ctypes.c_int(angle))

您的 C 代码可能如下所示:

unsigned int transform(unsigned char* a, int row, int col, int ang){
    for (int i = 0; i < 4; i++) {
        cout << (int)a[i] << endl; // add [0] to access the first row
    }
}

指针unsigned char* a不是指向二维数组的指针,而是指向一维数组(行主格式)的指针。


为了将二维 NumPy 数组传递给 C,您最好将 stride 参数添加到您的 C 函数。
步幅是在两个连续行之间步进的字节数。


为了测试,您可以填写每行的第 4 个元素:

real_image[0, 0] = 1
real_image[1, 0] = 2
real_image[2, 0] = 3
real_image[3, 0] = 4

您的 C 代码可能如下所示:

unsigned int transform(unsigned char* a, int row, int col, int stride, int ang){
    //Print the first element of each one of the first 4 rows
    for (int i = 0; i < 4; i++) {
        cout << (int)a[i*stride] << endl;
    }
}

注意新的 stride 参数,以及 a[i*stride]


正在从 Python 执行 transform:

hllDll.transform(n,ctypes.c_int(real_image.shape[0]), ctypes.c_int(real_image.shape[1]), ctypes.c_int(real_image.strides[0]), ctypes.c_int(angle))

注:

  • 很有可能 col = stride,但并非总是如此。