如何在Python 3中模拟4位整数？

Question

我想模拟无符号4位整数的溢出行为,如下所示:

>>> x, y = Int4(10), Int4(9)
>>> x + y
Int4(3)
>>> x * y
Int4(10)

内置的继承int似乎有效.是否可以在Int4不重写操作符方法的情况下实现类__add__？

Answer 1

不,子类化int在将算术应用于它时不会自动重用该类型:

>>> class Int4(int):
...     def __new__(cls, i):
...         return super(Int4, cls).__new__(cls, i & 0xf)
... 
>>> x, y = Int4(10), Int4(9)
>>> x + y
19
>>> type(x + y)
<type 'int'>

当你这样做时,你必须重写__add__等方法Int4().

如果您只想支持类型本身(例如,不支持在流程中转换其他数字类型),那么您可以生成大部分:

from functools import wraps

class Int4(int):
    def __new__(cls, i):
        return super(Int4, cls).__new__(cls, i & 0xf)

def add_special_method(cls, name):
    mname = '__{}__'.format(name)
    @wraps(getattr(cls, mname))
    def convert_to_cls(self, other):
        bound_original = getattr(super(cls, self), mname)
        return type(self)(bound_original(other))
    setattr(cls, mname, convert_to_cls)

for m in ('add', 'sub', 'mul', 'floordiv', 'mod', 'pow',
          'lshift', 'rshift', 'and', 'xor', 'or'):
    add_special_method(Int4, m)
    add_special_method(Int4, 'r' + m)  # reverse operation

这样生成的方法总是返回self算术特殊方法的类型; 这也允许进一步的子类化Int4.

演示:

>>> from functools import wraps
>>> class Int4(int):
...     def __new__(cls, i):
...         return super(Int4, cls).__new__(cls, i & 0xf)
... 
>>> def add_special_method(cls, name):
...     mname = '__{}__'.format(name)
...     @wraps(getattr(cls, mname))
...     def convert_to_cls(self, other):
...         bound_original = getattr(super(cls, self), mname)
...         return type(self)(bound_original(other))
...     setattr(cls, mname, convert_to_cls)
... 
>>> for m in ('add', 'sub', 'mul', 'floordiv', 'mod', 'pow',
...           'lshift', 'rshift', 'and', 'xor', 'or'):
...     add_special_method(Int4, m)
...     add_special_method(Int4, 'r' + m)  # reverse operation
... 
>>> x, y = Int4(10), Int4(9)
>>> x + y
3
>>> x * y
10