python中结构中的动态数组和结构

时间:2011-12-05 21:35:16

标签: python ctypes

我正在尝试使用ctypes在python中实现这个C结构:

struct _rows {
    int cols_count;
    char *cols[];
}

struct _unit {
    int rows_count;
    struct _rows *rows;
}

int my_func(struct _unit *param);

问题是_rows.cols是一个动态大小的char指针数组,_unit.rows是一个动态大小的_rows结构数组。如何在python中使用ctypes实现这一点?

我能够定义一个函数,它将返回一个带有可变数量的char指针的_rows结构:

def get_row(cols):
    class Row(ctypes.Structure):
        _fields_ = [("cols_count", ctypes.c_int), 
                    ("cols", ctypes.c_char_p * cols)
                   ]

我不知道该怎么做nex,它有点模糊,ctypes文档没有帮助。

1 个答案:

答案 0 :(得分:10)

我正在对OP想要的内容做出一些假设,如果有更简单的方法,我会喜欢建议,但这就是我想出来的:

demo.py

import string
from ctypes import Structure,c_int,c_char_p,POINTER,cast,pointer,byref,CDLL

class Row(Structure):
    _fields_ = [('cols_count', c_int), 
                ('cols', POINTER(c_char_p))]
    def __init__(self,cols):
        self.cols_count = cols
        # Allocate an array of character pointers
        pc = (c_char_p * cols)()
        self.cols = cast(pc,POINTER(c_char_p))            

class Unit(Structure):
    _fields_ = [('rows_count', c_int),
                ('rows',POINTER(Row))]
    def __init__(self,rows,cols):
        self.rows_count = rows
        # Allocate an array of Row structures.
        # This does NOT call __init__.
        pr = (Row * rows)()
        # Call init manually with the column size.
        for r in pr:
            r.__init__(cols)
        self.rows = cast(pr,POINTER(Row))

unit = Unit(2,3)

# Stuff some strings ('aaaaa','bbbbb',etc.)
for i in xrange(unit.rows_count):
    for j in xrange(unit.rows[i].cols_count):
        unit.rows[i].cols[j] = string.ascii_lowercase[i*5+j]*5

dll = CDLL('test.dll')
dll.my_func(byref(unit))

test.c的

#include <stdio.h>

struct _rows {
    int cols_count;
    char **cols;
};

struct _unit {
    int rows_count;
    struct _rows *rows;
};

__declspec(dllexport) int my_func(struct _unit *param)
{
    int i,j;
    for(i=0;i<param->rows_count;i++)
        for(j=0;j<param->rows[i].cols_count;j++)
            printf("%d,%d = %s\n",i,j,param->rows[i].cols[j]);
    return 0;
}

生成文件

使用Visual Studio 2010编译。

test.dll: test.c
    cl /W4 /LD test.c

输出

0,0 = aaaaa
0,1 = bbbbb
0,2 = ccccc
1,0 = fffff
1,1 = ggggg
1,2 = hhhhh