我有一个简单的类,它扩展 long 以接受带有值修饰符的字符串(即“10m”将是 1024*1024*10)
我有 __str__
函数打印传入的原始值(即如果传入“10m”,则返回“10m”)
问题是当我调用诸如:
>>> printf("%d" % Size('10m'))
我得到以下信息
SystemError: ../Objects/stringobject.c:4044: bad argument to internal function
显然,如果我打印 "%s"
,我会得到 '10m'
所以问题是,既然我是 long 的子类,为什么该类在应该获取 long 值时调用 __str__
。
顺便说一句,更多的测试表明 %x
和 %f
将打印整数值,这让我更加困惑。我也尝试添加 __format__
但它似乎只在调用 "...".format()
时被调用。
编辑 #1,这是代码:
class Size(long):
'''Represents a size reflected bytes. Subclass of long.
Size passed in must be in the formats <int> or "0x<int>" or "0x<int><unit>" or "<int><unit>" or "<int><unit><int><unit>....".
"0x<int><unit>0x<int><unit>" or similar numbers are not supported as is "<int><unit><int>"
b = bytes
s = sectors (512-byte)
k = kilobytes
m = megabytes
g = gigabytes
t = terabytes
'''
units = { 'b':1, 's':512, 'k':1024, 'm':1024 ** 2, 'g':1024 ** 3, 't':1024 ** 4 }
def __new__(cls, value):
'''Creates a Size object with the specified value.
Value can be a number or a string (optionally prefixed with '0x' or
postfixed with a type character). If using hex, the final character
will be treated as part of the value if it is a hex digit, regardless
of whether it is a valid unit character.
Examples:
Size(50)
Size("0x100s") # 256 sectors
Size("64")
Size("512k")
Size("0x1b") # this is 1b bytes, not 1 byte
'''
self = _new_unit_number(value, cls.units, long, cls)
return self
def __init__(self, value):
self._orig_value = value
def __str__(self):
print "calling str"
return str(self._orig_value) # Convert to str in case the object was created w/an int
def __format__(self, format_spec):
print "calling format"
print format_spec
try:
value = format(str(self), format_spec)
except ValueError:
value = format(int(self), format_spec)
return value
def _new_unit_number(value, unit_list, num_type, cls):
'''Converts a string of numbers followed by a unit character to the
requested numeric type (int or long for example).
'''
base = 10
start = 0
digits = string.digits
try:
if value[0:2] == '0x':
start = 2
base = 16
digits = string.hexdigits
if value[-1] in digits:
return num_type.__new__(cls, value[start:], base)
else:
try:
# Use a regex to split the parts of the unit
regex_string = '(\d+[%s])' % (''.join(unit_list.keys()))
parts = [x for x in re.split(regex_string, value[start:]) if x]
if len(parts) == 1:
return num_type.__new__(cls, num_type(value[start:-1], base) * unit_list[value[-1]])
else:
# Total up each part
# There's probably a better way to do this.
# This converts each unit to its base type, stores it in total,
# only to be converted back to the base type.
total = 0
for part in parts:
total += num_type(part[start:-1], base) * unit_list[part[-1]]
# Finally return the requested unit
return num_type.__new__(cls, total)
except KeyError:
raise ValueError("Invalid %s unit identifier: %s"
% (cls.__name__, unit_list[value[-1]]))
# not a string or empty, see if we can still use the class's constructor
except (TypeError, IndexError):
return num_type.__new__(cls, value)
最佳答案
不是真正的答案,但评论太长了。
我觉得这个问题非常有趣。我尝试使用此复制行为:
#! /usr/bin/python2.7
class Size (long):
def __new__ (cls, arg):
if arg and type (arg) == str:
if arg [-1] == 'm':
return super (Size, cls).__new__ (cls, long (arg [:-1] ) * 2 ** 20)
return super (Size, cls).__new__ (cls, arg)
def __init__ (self, arg):
self.s = arg
def __str__ (self):
return self.s
a = Size ('12m')
print (a)
print ('%s' % a)
#The following fails horribly
print ('%d' % a)
OP 描述的行为。但现在有趣的部分来了:当我从 int 而不是 long 继承时,它工作顺利:
class Size (int):
def __new__ (cls, arg):
if arg and type (arg) == str:
if arg [-1] == 'm':
return super (Size, cls).__new__ (cls, int (arg [:-1] ) * 2 ** 20)
return super (Size, cls).__new__ (cls, arg)
def __init__ (self, arg):
self.s = arg
def __str__ (self):
return self.s
也就是说,它在 python2 中工作正常,但在 python3 中失败。奇怪,奇怪。
关于python - 为什么 Python 调用 __str__ 而不是返回 long 值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20165983/