Question

我有一个包含以下条目的Python字典：

Tmp={'Name1': [10.0, 20.0, 'Title1', 1], 'Name2': [5.0, 25.0, 'Title2', 2]}

我想将它传递给C函数，函数定义为：

struct CA {
   char *Keys;
   float *Values;
   char *Title;
   int Index;
};

void myfunc (struct CA *in, int n);

在Python方面，我创建了一个等效的ctypes结构：

class CA(ctypes.Structure):
   _fields_ = [("Keys", ctypes.POINTER(ctypes.c_char_p)), 
               ("Values", ctypes.POINTER(ctypes.c_float)), 
               ("Title", ctypes.POINTER(ctypes.c_char_p)), 
               ("Index", ctypes.c_int)]

并使用以下方法创建了一个CA数组：

CAarray = CA * 2

现在我想在一个循环中将Tmp分配给CAarray，以便

k = Tmp.keys()
for (j, _) in enumerate(k):
   CAarray[j].Keys   = _ 
   CAarray[j].Values = Tmp[_][:2] 
   CAarray[j].Title  = Tmp[_][2]
   CAarray[j].Index  = Tmp[_][3]

我一直在努力使语法正确，并且到目前为止都失败了。帮助

另一方面，是否有任何例程/ lib可以处理Python变量和ctypes变量之间的转换？

Answer 1

我创建了一个测试DLL来验证结构是否正确传递。

#include <stdio.h>

struct CA {
   char *Keys;
   float *Values;
   char *Title;
   int Index;
};

__declspec(dllexport) void myfunc (struct CA *in, int n)
{
    int i;
    for(i = 0; i < n; ++i)
    {
        printf("%d: Keys = %s\n",i,in[i].Keys);
        printf("%d: Values = %f %f\n",i,in[i].Values[0],in[i].Values[1]);
        printf("%d: Title = %s\n",i,in[i].Title);
        printf("%d: Index = %d\n",i,in[i].Index);
    }
}

以下是我的称呼方式：

#!python3
from ctypes import *

class CA(Structure):
    _fields_ = [('Keys',c_char_p),
                ('Values',POINTER(c_float)),
                ('Title',c_char_p),
                ('Index',c_int)]

Tmp={'Name1': [10.0, 20.0, 'Title1', 1], 'Name2': [5.0, 25.0, 'Title2', 2]}

# repackage Tmp as a list of CA structures
ca_list = []
for k,v in Tmp.items():
    ca = CA()
    ca.Keys = k.encode('utf8') # Python 3 strings are Unicode, char* needs a byte string
    ca.Values = (c_float*2)(v[0],v[1]) # Interface unclear, how would target function know how many floats?
    ca.Title = v[2].encode('utf8')
    ca.Index = v[3]
    ca_list.append(ca)

# repackage python list of CAs to ctype array of CAs
ca_array = (CA * len(ca_list))(*ca_list)

dll = CDLL('test')
dll.myfunc.argtypes = POINTER(CA),c_int
dll.myfunc.restype = None

dll.myfunc(ca_array,len(ca_array))

输出：

0: Keys = Name1
0: Values = 10.000000 20.000000
0: Title = Title1
0: Index = 1
1: Keys = Name2
1: Values = 5.000000 25.000000
1: Title = Title2
1: Index = 2

Answer 2

不完全确定这一点，但似乎有效：

tmp = {'Name1': [10.0, 20.0, 'Title1', 1], 'Name2': [5.0, 25.0, 'Title2', 2]}

class CA(ctypes.Structure):
    _fields_ = [("key", ctypes.POINTER(ctypes.c_wchar_p)),
           ("values", ctypes.POINTER(ctypes.c_float)),
           ("title", ctypes.POINTER(ctypes.c_wchar_p)),
           ("index", ctypes.c_int)]

CAArray = CA * 2
ca_array = CAArray()

for ca, key in zip(ca_array, tmp):
    ca.key = ctypes.pointer(ctypes.c_wchar_p(key))
    ca.values = ctypes.pointer(ctypes.c_float(tmp[key][0]))
    ca.title = ctypes.pointer(ctypes.c_wchar_p(tmp[key][2]))
    ca.index = tmp[key][3]

for ca in ca_array:
    print(ca.key[0], ca.values[0], ca.values[1], ca.title[0], ca.index)

我坚持使用Python命名约定。

将Python字典转换为ctypes结构

2 个答案: