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)内建int的继承似乎有效。是否可以实现Int4类而不覆盖__add__之类的运算符方法?
发布于 2014-05-17 09:00:29
不,子类int在应用算术时不会自动重复使用该类型:
>>> 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'>您必须重写__add__等方法,以便在执行此操作时将其转换为Int4()。
如果您只想支持类型本身(例如,不支持转换进程中的其他数值类型),那么您可以生成其中的大多数:
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这将生成总是从算术特殊方法返回self类型的方法;这也将允许对Int4进行进一步的子类划分。
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发布于 2014-05-17 09:24:51
重写__add__方法是个好主意,因为您可以使计算看起来很清楚。Int4(4) + Int4(7)看起来比Int4(4).addTo(Int4(7))好(或者类似的东西)。下面是可以帮助您的代码:
class Int4:
def __init__(self, num): # initialising
self.num = self.cap(num)
def __str__(self):
return str(self.num)
def __repr__(self):
return "Int4(" + self.__str__() + ")"
def __add__(self, other): # addition
return Int4(self.cap(self.num + other.num))
def __sub__(self, other): # subtraction
return Int4(self.cap(self.num - other.num))
@staticmethod
def cap(num): # a method that handles an overflow
while num < 0:
num += 16
while num >= 16:
num -= 16
return num并测试:
>>> x,y,z = Int4(5), Int4(8), Int4(12)
>>> x
Int4(5)
>>> y
Int4(8)
>>> z
Int4(12)
>>> print x+y
13
>>> print z+y
4
>>> print x-z
9发布于 2015-06-11 21:46:24
这并不像@martijn的answer那样聪明,但它似乎适用于python2.7和3.*,而我在python2.7上得到了这个答案。
import sys
lt_py3 = sys.version_info < (3,)
lt_py33 = sys.version_info < (3, 3)
class Int(int):
'''
int types
'''
def __new__(self, val=0):
return int.__new__(self, val & (1 << self.bits - 1) - 1)
def __max_type_bits(self, other):
'''
determine the largest type and bits available from those in `self` and
`other`
'''
if hasattr(other, 'bits'):
if self.bits < other.bits:
return type(other), other.bits
return type(self), self.bits
def __unary_typed(oper):
'''
return a function that redefines the operation `oper` such that the
result conforms to the type of `self`
'''
def operate(self):
return type(self)(oper(self))
return operate
def __typed(oper):
'''
return a function that redefines the operation `oper` such that the
result conforms to the type of `self` or `other`, whichever is larger
if both are strongly typed (have a `bits` attribute); otherwise return
the result conforming to the type of `self`
'''
def operate(self, other):
typ, bits = self.__max_type_bits(other)
return typ(oper(self, other))
return operate
def __unary_ranged(oper):
'''
return a function that redefines the operator `oper` such that the
result conforms to both the range and the type of `self`
'''
def operate(self, other):
'''
type and bitmask the result to `self`
'''
return type(self)(oper(self) & (1 << self.bits - 1) - 1)
return operate
def __ranged(oper):
'''
return a function that redefines the operator `oper` such that the
result conforms to both the range and the type of `self` or `other`,
whichever is larger if both are strongly typed (have a `bits`
attribute); otherwise return the result conforming to the type of
`self`
'''
def operate(self, other):
'''
type and bitmask the result to either `self` or `other` whichever
is larger
'''
typ, bits = self.__max_type_bits(other)
return typ(oper(self, other) & (1 << bits - 1) - 1)
return operate
# bitwise operations
__lshift__ = __ranged(int.__lshift__)
__rlshift__ = __ranged(int.__rlshift__)
__rshift__ = __ranged(int.__rshift__)
__rrshift__ = __ranged(int.__rrshift__)
__and__ = __typed(int.__and__)
__rand__ = __typed(int.__rand__)
__or__ = __typed(int.__or__)
__ror__ = __typed(int.__ror__)
__xor__ = __typed(int.__xor__)
__rxor__ = __typed(int.__rxor__)
__invert__ = __unary_typed(int.__invert__)
# arithmetic operations
if not lt_py3:
__ceil__ = __unary_typed(int.__ceil__)
__floor__ = __unary_typed(int.__floor__)
__int__ = __unary_typed(int.__int__)
__abs__ = __unary_typed(int.__abs__)
__pos__ = __unary_typed(int.__pos__)
__neg__ = __unary_ranged(int.__neg__)
__add__ = __ranged(int.__add__)
__radd__ = __ranged(int.__radd__)
__sub__ = __ranged(int.__sub__)
__rsub__ = __ranged(int.__rsub__)
__mod__ = __ranged(int.__mod__)
__rmod__ = __ranged(int.__rmod__)
__mul__ = __ranged(int.__mul__)
__rmul__ = __ranged(int.__rmul__)
if lt_py3:
__div__ = __ranged(int.__div__)
__rdiv__ = __ranged(int.__rdiv__)
__floordiv__ = __ranged(int.__floordiv__)
__rfloordiv__ = __ranged(int.__rfloordiv__)
__pow__ = __ranged(int.__pow__)
__rpow__ = __ranged(int.__rpow__)
class Int4(Int):
bits = 4
x, y = Int4(10), Int4(9)
print(x + y)
print(x*y)在名为answer.py的文件中运行此代码将生成
$ python2.7 answer.py
3
2
$ python3.4 answer.py
3
2https://stackoverflow.com/questions/23709429
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