for循环双斯卡拉

我的代码是：for循环双斯卡拉

for(i <- -8.0 to 8.0 by 0.1) println(i)

输出是：

-8.0 
-7.9 
-7.800000000000001 
-7.700000000000001 
-7.600000000000001 
-7.500000000000002 
-7.400000000000002 
-7.3000000000000025 
-7.200000000000003 
-7.100000000000003 
-7.0000000000000036 
-6.900000000000004 
-6.800000000000004 
-6.700000000000005 
-6.600000000000005 
-6.500000000000005 
-6.400000000000006 
-6.300000000000006 
-6.200000000000006 
-6.100000000000007 
-6.000000000000007 
-5.9000000000000075 
-5.800000000000008 
-5.700000000000008 
-5.6000000000000085 
-5.500000000000009 
-5.400000000000009 
-5.30000000000001 
-5.20000000000001 
-5.10000000000001 
-5.000000000000011 
-4.900000000000011 
-4.800000000000011 
-4.700000000000012 
-4.600000000000012 
-4.500000000000012 
-4.400000000000013 
-4.300000000000013 
-4.2000000000000135 
-4.100000000000014 
-4.000000000000014 
-3.900000000000014 
-3.800000000000014 
-3.700000000000014 
-3.600000000000014 
-3.5000000000000138 
-3.4000000000000137 
-3.3000000000000136 
-3.2000000000000135 
-3.1000000000000134 
-3.0000000000000133 
-2.9000000000000132 
-2.800000000000013 
-2.700000000000013 
-2.600000000000013 
-2.500000000000013 
-2.400000000000013 
-2.3000000000000127 
-2.2000000000000126 
-2.1000000000000125 
-2.0000000000000124 
-1.900000000000
-1.800000000000
-1.7000000000000122 
-1.600000000000012 
-1.500000000000012 
-1.400000000000012 
-1.3000000000000118 
-1.2000000000000117 
-1.1000000000000116 
-1.0000000000000115 
-0.9000000000000116 
-0.8000000000000116 
-0.7000000000000116 
-0.6000000000000116 
-0.5000000000000117 
-0.4000000000000117 
-0.3000000000000117 
-0.2000000000000117 
-0.10000000000001169 
-1.1685097334179773E-14 
0.09999999999998832 
0.19999999999998833 
0.29999999999998833 
0.39999999999998836 
0.49999999999998834 
0.5999999999999883 
0.6999999999999883 
0.7999999999999883 
0.8999999999999883 
0.9999999999999882 
1.0999999999999883 
1.1999999999999884 
1.2999999999999885 
1.3999999999999886 
1.4999999999999887 
1.5999999999999888 
1.6999999999999889 
1.799999999999989 
1.899999999999989 
1.9999999999999891 
2.099999999999989 
2.199999999999989 
2.299999999999989 
2.3999999999999893 
2.4999999999999893 
2.5999999999999894 
2.6999999999999895 
2.7999999999999896 
2.8999999999999897 
2.99999999999999 
3.09999999999999 
3.19999999999999 
3.29999999999999 
3.39999999999999 
3.4999999999999902 
3.5999999999999903 
3.6999999999999904 
3.7999999999999905 
3.8999999999999906 
3.9999999999999907 
4.099999999999991 
4.19999999999999 
4.29999999999999 
4.39999999999999 
4.499999999999989 
4.599999999999989 
4.699999999999989 
4.799999999999988 
4.899999999999988 
4.999999999999988 
5.099999999999987 
5.199999999999987 
5.2999999999999865 
5.399999999999986 
5.499999999999986 
5.599999999999985 
5.699999999999985 
5.799999999999985 
5.899999999999984 
5.999999999999984 
6.099999999999984 
6.199999999999983 
6.299999999999983 
6.399999999999983 
6.499999999999982 
6.599999999999982 
6.6999999999999815 
6.799999999999981 
6.899999999999981 
6.9999999999999805 
7.09999999999998 
7.19999999999998 
7.299999999999979 
7.399999999999979 
7.499999999999979 
7.599999999999978 
7.699999999999978 
7.799999999999978 
7.899999999999977 
7.999999999999977

我希望得到的结果为：

(-8.0, -7.9, -7.8, -7.7......., 7.7, 7.8, 7.9, 8.0)

我应该如何解决这个问题？

来源

2017-01-18 Gaurav Garg

你无法修复它。 https://en.wikipedia.org/wiki/Floating_point#Accuracy_problems – Reactormonk

我欣赏接受:-) – GhostCat

你的期望是错的。你认为浮点数在计算机上的工作方式与“纯数学”中的相同。

那么，他们不会，看到here。

所以，你要明白，你的循环构建自身

for(i <- -8.0 to 8.0 by 0.1)

，你期望它不起作用。

来源

2017-01-18 13:41:32 GhostCat

我不知道我同意这只是一个字符串格式问题。 –

但事实并非如此。这是关于浮点的固有不准确性。 –

“你需要什么：舍入/ resp。格式化你的double值。”和“为了得到预期的输出，你必须把所有的值变成字符串”是错误的。继续做一些本质上不准确的事情，并且在产出中进行篡改并不是正确的方式（显然，IMO）。根据确切的值，像'for（i <-8.0到8.0 by 0.1）'可能会产生意想不到的元素数量（因为预期的最终测试由于不准确而失败），并且舍入/格式化/将double值转换为字符串将会解决这个问题。 –

for循环双斯卡拉

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