是否有一种方法可以方便地在Python中定义类c结构?我厌倦了写这样的东西:

class MyStruct():
    def __init__(self, field1, field2, field3):
        self.field1 = field1
        self.field2 = field2
        self.field3 = field3

当前回答

有一个python包正是用于此目的。看到cstruct2py

cstruct2py是一个纯python库,用于从C代码生成python类,并使用它们来打包和解包数据。该库可以解析C头文件(结构体、联合、枚举和数组声明),并在python中模拟它们。生成的python类可以解析和打包数据。

例如:

typedef struct {
  int x;
  int y;
} Point;

after generating pythonic class...
p = Point(x=0x1234, y=0x5678)
p.packed == "\x34\x12\x00\x00\x78\x56\x00\x00"

如何使用

首先,我们需要生成python结构体:

import cstruct2py
parser = cstruct2py.c2py.Parser()
parser.parse_file('examples/example.h')

现在我们可以从C代码中导入所有的名称:

parser.update_globals(globals())

我们也可以直接这样做:

A = parser.parse_string('struct A { int x; int y;};')

使用C代码中的类型和定义

a = A()
a.x = 45
print a
buf = a.packed
b = A(buf)
print b
c = A('aaaa11112222', 2)
print c
print repr(c)

输出将是:

{'x':0x2d, 'y':0x0}
{'x':0x2d, 'y':0x0}
{'x':0x31316161, 'y':0x32323131}
A('aa111122', x=0x31316161, y=0x32323131)

克隆

对于克隆cstruct2py运行:

git clone https://github.com/st0ky/cstruct2py.git --recursive

其他回答

这里的一些答案非常详尽。我找到的最简单的选项是(from: http://norvig.com/python-iaq.html):)

class Struct:
    "A structure that can have any fields defined."
    def __init__(self, **entries): self.__dict__.update(entries)

初始化:

>>> options = Struct(answer=42, linelen=80, font='courier')
>>> options.answer
42

添加更多的:

>>> options.cat = "dog"
>>> options.cat
dog

编辑:对不起,没有看到这个例子已经进一步。

一本字典怎么样?

就像这样:

myStruct = {'field1': 'some val', 'field2': 'some val'}

然后你可以使用这个来操作值:

print myStruct['field1']
myStruct['field2'] = 'some other values'

值不一定是字符串。它们可以是其他任何物体。

如果您没有3.7的@dataclass,并且需要可变性,那么下面的代码可能适合您。它是非常自文档化和ide友好的(自动完成),防止编写两次内容,易于扩展,并且非常简单地测试所有实例变量都被完全初始化:

class Params():
    def __init__(self):
        self.var1 : int = None
        self.var2 : str = None

    def are_all_defined(self):
        for key, value in self.__dict__.items():
            assert (value is not None), "instance variable {} is still None".format(key)
        return True


params = Params()
params.var1 = 2
params.var2 = 'hello'
assert(params.are_all_defined)

下面结构的解决方案是受namedtuple实现和前面一些答案的启发。然而,与namedtuple不同的是,它的值是可变的,但就像c风格的结构体在名称/属性中是不可变的,而普通的类或dict不是。

_class_template = """\
class {typename}:
def __init__(self, *args, **kwargs):
    fields = {field_names!r}

    for x in fields:
        setattr(self, x, None)            

    for name, value in zip(fields, args):
        setattr(self, name, value)

    for name, value in kwargs.items():
        setattr(self, name, value)            

def __repr__(self):
    return str(vars(self))

def __setattr__(self, name, value):
    if name not in {field_names!r}:
        raise KeyError("invalid name: %s" % name)
    object.__setattr__(self, name, value)            
"""

def struct(typename, field_names):

    class_definition = _class_template.format(
        typename = typename,
        field_names = field_names)

    namespace = dict(__name__='struct_%s' % typename)
    exec(class_definition, namespace)
    result = namespace[typename]
    result._source = class_definition

    return result

用法:

Person = struct('Person', ['firstname','lastname'])
generic = Person()
michael = Person('Michael')
jones = Person(lastname = 'Jones')


In [168]: michael.middlename = 'ben'
Traceback (most recent call last):

  File "<ipython-input-168-b31c393c0d67>", line 1, in <module>
michael.middlename = 'ben'

  File "<string>", line 19, in __setattr__

KeyError: 'invalid name: middlename'

https://stackoverflow.com/a/32448434/159695在Python3中不起作用。

https://stackoverflow.com/a/35993/159695在Python3中工作。

然后我扩展它来添加默认值。

class myStruct:
    def __init__(self, **kwds):
        self.x=0
        self.__dict__.update(kwds) # Must be last to accept assigned member variable.
    def __repr__(self):
        args = ['%s=%s' % (k, repr(v)) for (k,v) in vars(self).items()]
        return '%s(%s)' % ( self.__class__.__qualname__, ', '.join(args) )

a=myStruct()
b=myStruct(x=3,y='test')
c=myStruct(x='str')

>>> a
myStruct(x=0)
>>> b
myStruct(x=3, y='test')
>>> c
myStruct(x='str')