这可能是一个迟来的答案，但我终于让它工作了。所有的功劳都归sturla molden在this link。
关键是，注意double**是np.uintp类型的数组。

xpp = (x.ctypes.data + np.arange(x.shape[0]) * x.strides[0]).astype(np.uintp)
doublepp = np.ctypeslib.ndpointer(dtype=np.uintp)

然后使用doublepp做为型别，传入xpp。请参阅随附的完整程式码。
C代码：

// dummy.c 
#include <stdlib.h> 

__declspec(dllexport) void foobar(const int m, const int n, const 
double **x, double **y) 
{ 
    size_t i, j; 
    for(i=0; i<m; i++) 
        for(j=0; j<n; j++) 
            y[i][j] = x[i][j]; 
}

Python代码：

# test.py 
import numpy as np 
from numpy.ctypeslib import ndpointer 
import ctypes 

_doublepp = ndpointer(dtype=np.uintp, ndim=1, flags='C') 

_dll = ctypes.CDLL('dummy.dll') 

_foobar = _dll.foobar 
_foobar.argtypes = [ctypes.c_int, ctypes.c_int, _doublepp, _doublepp] 
_foobar.restype = None 

def foobar(x): 
    y = np.zeros_like(x) 
    xpp = (x.__array_interface__['data'][0] 
      + np.arange(x.shape[0])*x.strides[0]).astype(np.uintp) 
    ypp = (y.__array_interface__['data'][0] 
      + np.arange(y.shape[0])*y.strides[0]).astype(np.uintp) 
    m = ctypes.c_int(x.shape[0]) 
    n = ctypes.c_int(x.shape[1]) 
    _foobar(m, n, xpp, ypp) 
    return y 

if __name__ == '__main__': 
    x = np.arange(9.).reshape((3, 3)) 
    y = foobar(x)

希望能有所帮助，
尚

#include <stdio.h>

void test(double (*in_array)[3], int N){
    int i, j;

    for(i = 0; i < N; i++){
        for(j = 0; j < N; j++){
            printf("%e \t", in_array[i][j]);
        }
        printf("\n");
    }
}

int main(void)
{
    double a[][3] = {
        {1., 2., 3.},
        {4., 5., 6.},
        {7., 8., 9.},
    };

    test(a, 3);
    return 0;
}

如果要在函数中使用double **，必须将指针数组传递给double（而不是2d数组）：

#include <stdio.h>

void test(double **in_array, int N){
    int i, j;

    for(i = 0; i < N; i++){
        for(j = 0; j< N; j++){
            printf("%e \t", in_array[i][j]);
        }
        printf("\n");
    }
}

int main(void)
{
    double a[][3] = {
        {1., 2., 3.},
        {4., 5., 6.},
        {7., 8., 9.},
    };
    double *p[] = {a[0], a[1], a[2]};

    test(p, 3);
    return 0;
}

另一个（如@eryksun所建议）：传递一个指针并进行一些运算以获得索引：

#include <stdio.h>

void test(double *in_array, int N){
    int i, j;

    for(i = 0; i < N; i++){
        for(j = 0; j< N; j++){
            printf("%e \t", in_array[i * N + j]);
        }
        printf("\n");
    }
}

int main(void)
{
    double a[][3] = {
        {1., 2., 3.},
        {4., 5., 6.},
        {7., 8., 9.},
    };

    test(a[0], 3);
    return 0;
}

虽然答复可能较迟，但我希望它可以帮助其他有同样问题的人。
由于numpy数组在内部保存为1d数组，因此可以在C中简单地重建2d形状。

// libtest2d.c
#include <stdlib.h> // for malloc and free
#include <stdio.h>  // for printf

// create a 2d array from the 1d one
double ** convert2d(unsigned long len1, unsigned long len2, double * arr) {
    double ** ret_arr;

    // allocate the additional memory for the additional pointers
    ret_arr = (double **)malloc(sizeof(double*)*len1);

    // set the pointers to the correct address within the array
    for (int i = 0; i < len1; i++) {
        ret_arr[i] = &arr[i*len2];
    }

    // return the 2d-array
    return ret_arr;
}

// print the 2d array
void print_2d_list(unsigned long len1,
    unsigned long len2,
    double * list) {

    // call the 1d-to-2d-conversion function
    double ** list2d = convert2d(len1, len2, list);

    // print the array just to show it works
    for (unsigned long index1 = 0; index1 < len1; index1++) {
        for (unsigned long index2 = 0; index2 < len2; index2++) {
            printf("%1.1f ", list2d[index1][index2]);
        }
        printf("\n");
    }

    // free the pointers (only)
    free(list2d);
}

和

# test2d.py

import ctypes as ct
import numpy as np

libtest2d = ct.cdll.LoadLibrary("./libtest2d.so")
libtest2d.print_2d_list.argtypes = (ct.c_ulong, ct.c_ulong,
        np.ctypeslib.ndpointer(dtype=np.float64,
            ndim=2,
            flags='C_CONTIGUOUS'
            )
        )
libtest2d.print_2d_list.restype = None

arr2d = np.meshgrid(np.linspace(0, 1, 6), np.linspace(0, 1, 11))[0]

libtest2d.print_2d_list(arr2d.shape[0], arr2d.shape[1], arr2d)

如果您使用gcc -shared -fPIC libtest2d.c -o libtest2d.so编译代码，然后运行python test2d.py，它应该打印数组。
我希望这个例子多少能说明一些问题。其思想是，形状也被赋予C代码，然后C代码创建一个double **指针，为该指针保留了额外指针的空间。然后这些指针被设置为指向原始数组的正确部分。
PS：我是一个相当初学者在C，所以请评论，如果有理由不这样做。

这里我传递了两个二维numpy数组，并打印了一个数组的值作为引用

您可以在cpp中使用和编写自己的逻辑
cpp_function.cpp
编译时使用：g++ -shared -fPIC cpp_function.cpp -o cpp_function.so

#include <iostream>
extern "C" {
void mult_matrix(double *a1, double *a2, size_t a1_h, size_t a1_w, 
                  size_t a2_h, size_t a2_w, int size)
{
    //std::cout << "a1_h & a1_w" << a1_h << a1_w << std::endl; 
    //std::cout << "a2_h & a2_w" << a2_h << a2_w << std::endl; 
    for (size_t i = 0; i < a1_h; i++) {
        for (size_t j = 0; j < a1_w; j++) {
            printf("%f ", a1[i * a1_h + j]);
        }
        printf("\n");
    }
    printf("\n");
  }

}

Python文件main.py

import ctypes
import numpy
from time import time

libmatmult = ctypes.CDLL("./cpp_function.so")
ND_POINTER_1 = numpy.ctypeslib.ndpointer(dtype=numpy.float64, 
                                      ndim=2,
                                      flags="C")
ND_POINTER_2 = numpy.ctypeslib.ndpointer(dtype=numpy.float64, 
                                    ndim=2,
                                    flags="C")
libmatmult.mult_matrix.argtypes = [ND_POINTER_1, ND_POINTER_2, ctypes.c_size_t, ctypes.c_size_t]
# print("-->", ctypes.c_size_t)

def mult_matrix_cpp(a,b):
    shape = a.shape[0] * a.shape[1]
    libmatmult.mult_matrix.restype = None
    libmatmult.mult_matrix(a, b, *a.shape, *b.shape , a.shape[0] * a.shape[1])

size_a = (300,300)
size_b = size_a

a = numpy.random.uniform(low=1, high=255, size=size_a)
b = numpy.random.uniform(low=1, high=255, size=size_b)

t2 = time()
out_cpp = mult_matrix_cpp(a,b)
print("cpp time taken:{:.2f} ms".format((time() - t2) * 1000))
out_cpp = numpy.array(out_cpp).reshape(size_a[0], size_a[1])