小编Jus*_*ter的帖子

我刚刚开始使用python,我想知道如何用python定义联合(使用ctypes)？希望我通过ctypes支持工会是对的.例如,以下c代码是如何在python中的

struct test
{
char something[10];
int status;
};

struct test2
{
char else[10];
int status;
int alive;
};

union tests
{
struct test a;
struct test2 b;
};

struct tester
{
char more_chars[20];
int magic;
union tests f;
};

Thx,如果其他人正在寻找相同的答案,则添加简单示例

from ctypes import *

class POINT(Structure):
    _fields_ = [("x", c_int),
                 ("y", c_int)]

class POINT_1(Structure):
    _fields_ = [("x", c_int),
                 ("y", c_int),
                 ("z",c_int)]

class POINT_UNION(Union):
    _fields_ = [("a", POINT),
                 ("b", POINT_1)]

class TEST(Structure):
    _fields_ = [("magic", c_int),
                 ("my_union", POINT_UNION)]

testing = …

我有以后的代码,它最终导致分段错误.

import ctypes
from random import randint

class STRUCT_2(ctypes.Structure):
    #_pack_=2
    _fields_ = [('field_1', ctypes.c_short),
                ('field_2', ctypes.c_short),
                ('field_3', ctypes.c_short)]

class STRUCT_1(ctypes.Structure):
    #_pack_=2
    _fields_ = [('elements', ctypes.c_short),
                ('STRUCT_ARRAY', ctypes.POINTER(STRUCT_2))]

    def __init__(self,num_of_structs):
        elems = (ctypes.POINTER(STRUCT_2) * num_of_structs)()
        self.STRUCT_ARRAY = ctypes.cast(elems,ctypes.POINTER(STRUCT_2))
        self.elements = num_of_structs

        for num in range(0,num_of_structs):
            self.STRUCT_ARRAY[num].field_1 = 1
            self.STRUCT_ARRAY[num].field_2 = 2
            self.STRUCT_ARRAY[num].field_3 = 3

for num in range(0,100):
    test = STRUCT_1(num)
    print "%i done" % num

输出:5完成分段故障

但是如果struct_2中没有field_3那么它似乎正常工作.如果我添加一个短字段(field_4),它会结束分段错误...

那么我做错了什么或我错过了什么？

还有其他方法来定义数组大小吗？

所以我有两个简单的 ctypes 结构

class S2 (ctypes.Structure):
    _fields_ = [
    ('A2',     ctypes.c_uint16*10),
    ('B2',     ctypes.c_uint32*10),
    ('C2',     ctypes.c_uint32*10) ]


class S1 (ctypes.Structure):
    _fields_ = [
    ('A',     ctypes.c_uint16),
    ('B',     ctypes.c_uint32),
    ('C',     S2) ]

例如，是否可以使用namedtuple 执行相同的操作？nametuple 中如何处理列表？

编辑：

结构包用法

test_data = '0100000002000000' + 10*'01' + 10*'01' + 10*'01'

S2 = collections.namedtuple('S2', ['A2', 'B2', 'C2'])
S1 = collections.namedtuple('S1', ['A', 'B', 'REF_to_S2'])

Data2 = S2._make(struct.unpack('10p10p10p', binascii.unhexlify(test_data[16:])))
##this is not working, because there must be 3 args..
Data1 = S1._make(struct.unpack('ii', binascii.unhexlify(test_data[0:16])))

最后我想以可读格式打印数据（具有可见的键：值对）。但现在我不知道应该如何处理两个不同的命名元组的解包操作......？

这个 unpack.struct 操作会处理值类型问题，对吗？