Java 使用 Math.round 四舍五入到最接近的十分之一？
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Using Math.round to round to the nearest tenth?
提问by Travis M
I just wanted to ask a quick question regarding the Math.round method. I'm trying to compute the division of two ints into a double. The equation looks like this: 199/39. When I do this it returns 5.0 as the answer. I know the answer should be 5.1 with some more decimals. I have been told to use the Math.round method to round it to the nearest tenth, but I have no idea how to accomplish this. Should I change that double variable to a int and make it int/int=int? I'm not sure how Math.round even works to get 5.1 as I've read it only rounds to the nearest integer not decimal point. Any help would be fantastic.
我只是想问一个关于 Math.round 方法的快速问题。我正在尝试将两个整数的除法计算为双精度。等式如下所示：199/39。当我这样做时，它返回 5.0 作为答案。我知道答案应该是 5.1，再加上一些小数。有人告诉我使用 Math.round 方法将其四舍五入到最接近的十分之一，但我不知道如何实现这一点。我应该将该双变量更改为 int 并使其成为 int/int=int 吗？我不确定 Math.round 是如何获得 5.1 的，因为我读过它只四舍五入到最接近的整数而不是小数点。任何帮助都会很棒。
P.S This is homework, but I ask only because I can't find any information in my notes, slides, or book on how to use Math.round.
PS 这是作业，但我问这个只是因为我在我的笔记、幻灯片或书中找不到任何关于如何使用 Math.round 的信息。
回答by Michael Yaworski
Your specific answer:
您的具体回答：
roundedNumber = (double)Math.round(unRoundedNumber * 10) / 10;
In general, the equation is:
一般来说，等式是：
roundedNumber =
(double)Math.round(unRoundedNumber * Math.pow(10, digitsToRoundTo))
/ Math.pow(10, digitsToRoundTo);
回答by hexacyanide
You don't need Math.round()to get a resultant decimal value. If you divide an intby an int, you will get an int. if you want a decimal, then cast doubleto one of the input values. Then you will get a doubleas a result.
您不需要Math.round()获得结果的十进制值。如果将 anint除以 an int，您将得到 an int。如果您想要小数，则double转换为输入值之一。然后你会得到double一个结果。
(double) 199 / 39
199.0 / 39
// both return
5.102564102564102
回答by user207421
I know the answer should be 5.1 with some more decimals.
我知道答案应该是 5.1，再加上一些小数。
Not with integer division it shouldn't. 5 is correct.
不是整数除法，它不应该。5是正确的。
If you want a floating-point answer, you need to provide at least one floating-point operand, e.g. 199/39.0.
如果你想要一个浮点答案，你需要提供至少一个浮点操作数，例如199/39.0.
You can then format that for printing with as many or few decimal places you like, with System.printf()or DecimalFormat.
然后，您可以格式化它以使用您喜欢的任意多或少的小数位进行打印，使用System.printf()或DecimalFormat.
You can't round the floating-point value itself to decimal places, because it doesn't have decimal places, it has binary places.
您不能将浮点值本身四舍五入到小数位，因为它没有小数位，它有二进制位。
See this questionfor a full discussion, especially my answer there.
请参阅此问题以进行全面讨论，尤其是我在那里的回答。
回答by Travis M
Thanks for all the help. After reading your posts and looking at it some more I ended up doing ratio=((double)199/39) and then going ratio=(double)Math.round(ratio*10)/10. Doing that got me the 5.1 i was looking for.
感谢所有的帮助。在阅读了您的帖子并查看了更多内容后，我最终执行了 ratio=((double)199/39) 然后执行 ratio=(double)Math.round(ratio*10)/10。这样做让我得到了我正在寻找的 5.1。
