感谢您的回复。大多数人都很好，并提供了良好的链接，所以我就这样说，回答我自己的问题。
Caspin发布了此链接。
他还提到Google Tests使用ULP比较，当我查看google代码时，我发现它们提到了与cygnus软件相同的确切链接。
我结束了在C中将某些算法实现为Python扩展，然后发现我也可以在纯Python中实现它。该代码发布在下面。
最后，我可能只会在我的技巧包中添加ULP差异。
有趣的是，在两个永远不会离开内存的相等数字之间应该有多少个浮点数。我读过的一篇文章或Google代码说4是个不错的数字…但是在这里我能达到10。
>>> f1 = 25.4
>>> f2 = f1
>>>
>>> for i in xrange(1, 11):
... f2 /= 10.0 # to cm
... f2 *= (1.0 / 2.54) # to in
... f2 *= 25.4 # back to mm
... print 'after %2d loops there are %2d doubles between them' % (i, dulpdiff(f1, f2))
...
after 1 loops there are 1 doubles between them
after 2 loops there are 2 doubles between them
after 3 loops there are 3 doubles between them
after 4 loops there are 4 doubles between them
after 5 loops there are 6 doubles between them
after 6 loops there are 7 doubles between them
after 7 loops there are 8 doubles between them
after 8 loops there are 10 doubles between them
after 9 loops there are 10 doubles between them
after 10 loops there are 10 doubles between them
同样有趣的是，当其中一个作为字符串写出并读回时，相等的数字之间有多少个浮点。
>>> # 0 degrees Fahrenheit is -32 / 1.8 degrees Celsius
... f = -32 / 1.8
>>> s = str(f)
>>> s
'-17.7777777778'
>>> # floats between them...
... fulpdiff(f, float(s))
0
>>> # doubles between them...
... dulpdiff(f, float(s))
6255L
import struct
from functools import partial
# (c) 2010 Eric L. Frederich
#
# Python implementation of algorithms detailed here...
# from http://www.cygnus-software.com/papers/comparingfloats/comparingfloats.htm
def c_mem_cast(x, f=None, t=None):
'''
do a c-style memory cast
In Python...
x = 12.34
y = c_mem_cast(x, 'd', 'l')
... should be equivilent to the following in c...
double x = 12.34;
long y = *(long*)&x;
'''
return struct.unpack(t, struct.pack(f, x))[0]
dbl_to_lng = partial(c_mem_cast, f='d', t='l')
lng_to_dbl = partial(c_mem_cast, f='l', t='d')
flt_to_int = partial(c_mem_cast, f='f', t='i')
int_to_flt = partial(c_mem_cast, f='i', t='f')
def ulp_diff_maker(converter, negative_zero):
'''
Getting the ulp difference of floats and doubles is similar.
Only difference if the offset and converter.
'''
def the_diff(a, b):
# Make a integer lexicographically ordered as a twos-complement int
ai = converter(a)
if ai < 0:
ai = negative_zero - ai
# Make b integer lexicographically ordered as a twos-complement int
bi = converter(b)
if bi < 0:
bi = negative_zero - bi
return abs(ai - bi)
return the_diff
# double ULP difference
dulpdiff = ulp_diff_maker(dbl_to_lng, 0x8000000000000000)
# float ULP difference
fulpdiff = ulp_diff_maker(flt_to_int, 0x80000000 )
# default to double ULP difference
ulpdiff = dulpdiff
ulpdiff.__doc__ = '''
Get the number of doubles between two doubles.
'''
602392714
清零编程群