for循環雙斯卡拉

我的代碼是：for循環雙斯卡拉

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

輸出是：

-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

我希望得到的結果爲：

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

我應該如何解決這個問題？

來源

2017-01-18 Gaurav Garg

你無法修復它。 https://en.wikipedia.org/wiki/Floating_point#Accuracy_problems – Reactormonk

我欣賞接受:-) – GhostCat

你的期望是錯的。你認爲浮點數在計算機上的工作方式與「純數學」中的相同。

那麼，他們不會，看到here。

所以，你要明白，你的循環構建自身

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

，你期望它不起作用。

來源

2017-01-18 13:41:32 GhostCat

我不知道我同意這只是一個字符串格式問題。 –

但事實並非如此。這是關於浮點的固有不準確性。 –

「你需要什麼：舍入/ resp。格式化你的double值。」和「爲了得到預期的輸出，你必須把所有的值變成字符串」是錯誤的。繼續做一些本質上不準確的事情，並且在產出中進行篡改並不是正確的方式（顯然，IMO）。根據確切的值，像'for（i <-8.0到8.0 by 0.1）'可能會產生意想不到的元素數量（因爲預期的最終測試由於不準確而失敗），並且舍入/格式化/將double值轉換爲字符串將會解決這個問題。 –

for循環雙斯卡拉

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