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Fix a bug where comparison of a rational with a float failed because 

the difference got converted to float.
Put brackets around the string representation of (non-integer)
rationals.
(Sjoerd Mullender.)
This commit is contained in:
Guido van Rossum 1998-09-09 14:07:06 +00:00
parent 76d1f96fe2
commit 7ca9a1a466
1 changed files with 9 additions and 9 deletions
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18
Demo/classes/Rat.py
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@ -100,7 +100,7 @@ class Rat:
if self.__den == 1:
return str(self.__num)
else:
return '%s/%s' % (str(self.__num), str(self.__den))
return '(%s/%s)' % (str(self.__num), str(self.__den))
# a + b
def __add__(a, b):
@ -213,7 +213,7 @@ class Rat:
# cmp(a,b)
def __cmp__(a, b):
diff = a - b
diff = Rat(a - b)
if diff.__num < 0:
return -1
elif diff.__num > 0:
@ -244,16 +244,16 @@ def test():
[Rat(1,2), Rat(-3,10), Rat(1,25), Rat(1,4)]
[Rat(-3,10), Rat(1,25), Rat(1,4), Rat(1,2)]
0
11/10
11/10
(11/10)
(11/10)
1.1
OK
2 1.5 3/2 (1.5+1.5j) 15707963/5000000
2 1.5 (3/2) (1.5+1.5j) (15707963/5000000)
2 2 2.0 (2+0j)
4 0 4 1 4 0
3.5 0.5 3.0 1.33333333333 2.82842712475 1
7/2 1/2 3 4/3 2.82842712475 1
(7/2) (1/2) 3 (4/3) 2.82842712475 1
(3.5+1.5j) (0.5-1.5j) (3+3j) (0.666666666667-0.666666666667j) (1.43248815986+2.43884761145j) 1
1.5 1 1.5 (1.5+0j)
@ -261,11 +261,11 @@ def test():
3.0 0.0 2.25 1.0 1.83711730709 0
3.0 0.0 2.25 1.0 1.83711730709 1
(3+1.5j) -1.5j (2.25+2.25j) (0.5-0.5j) (1.50768393746+1.04970907623j) -1
3/2 1 1.5 (1.5+0j)
(3/2) 1 1.5 (1.5+0j)
7/2 -1/2 3 3/4 9/4 -1
(7/2) (-1/2) 3 (3/4) (9/4) -1
3.0 0.0 2.25 1.0 1.83711730709 -1
3 0 9/4 1 1.83711730709 0
3 0 (9/4) 1 1.83711730709 0
(3+1.5j) -1.5j (2.25+2.25j) (0.5-0.5j) (1.50768393746+1.04970907623j) -1
(1.5+1.5j) (1.5+1.5j)