将Kotlin的Double舍入到小数点后一位:从0.044999到0.1 [英] Round Double to 1 decimal place kotlin: from 0.044999 to 0.1

问题描述

我有一个Double变量0.0449999，我想将其四舍五入到小数点后一位0.1.

I have a Double variable that is 0.0449999 and I would like to round it to 1 decimal place 0.1 .

我正在使用Kotlin，但是Java解决方案也很有帮助.

I am using Kotlin but the Java solution is also helpful.

val number:Double = 0.0449999

我尝试使用以下两种解决方案将小数点后一位:

I tried getting 1 decimal place with these two solutions:

1. val solution = Math.round(number * 10.0) / 10.0
2. val solution = String.format("%.1f", number)

1. val solution = Math.round(number * 10.0) / 10.0
2. val solution = String.format("%.1f", number)

问题是两种情况下我都得到0.0，因为它会将数字从0.04舍入到0.0.并不需要将所有的小数点都取整.

The problem is that I get 0.0 in both cases because it rounds the number from 0.04 to 0.0. It doesn't take all decimals and round it.

我想获得0.1:0.045 -> 0.05 -> 0.1

推荐答案

最后，我按照Andy Turner的建议进行了操作，四舍五入到小数点后三位，然后四舍五入到1，然后再四舍五入:

Finally I did what Andy Turner suggested, rounded to 3 decimals, then to 2 and then to 1:

答案1:

val number:Double = 0.0449999
val number3digits:Double = String.format("%.3f", number).toDouble()
val number2digits:Double = String.format("%.2f", number3digits).toDouble()
val solution:Double = String.format("%.1f", number2digits).toDouble()

答案2:

val number:Double = 0.0449999
val number3digits:Double = Math.round(number * 1000.0) / 1000.0
val number2digits:Double = Math.round(number3digits * 100.0) / 100.0
val solution:Double = Math.round(number2digits * 10.0) / 10.0

结果:

0.045→0.05→0.1

0.045 → 0.05 → 0.1

注意:我知道这不是应该如何工作的，但是有时在某些特殊情况下，您需要考虑所有小数点后四舍五入，因此也许有人觉得这很有用.

Note: I know it is not how it should work but sometimes you need to round up taking into account all decimals for some special cases so maybe someone finds this useful.

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