甚至理解一半？ [英] Understanding round half even?

问题描述

 （new BigDecimal（5）* new BigDecimal（0.049））
 
 code> 

它给出了



  0.24500000000000000943689570931383059360086917877197265625 
  

当我使用ROUND_HALF_EVEN

 （new BigDecimal（5）* new BigDecimal（0.049））。setScale（2，BigDecimal.ROUND_HALF_EVEN）
  

它打印

  0.25 
  

所以我的困惑是，不应该是偶数，所以0.25不应该是0.24。请帮助解决这个问题。谢谢！

解决方案

这里的REAL问题是，您对 BigDecimal

可以做你想做的事。



问题在于 0.049 是浮点数字，并且该值不能完全表示为浮点数值（既不是 float 也不是 double ），引入了一个在这种情况下有意义的小错误。



通过使用接受 String 参数的构造函数，您可以避免通过浮点进行转换并获得确切的值打算。



计算机上的浮点算法由于其精度有限而充满了令人讨厌的意外行为。如果您想了解更多有关陷阱的信息，请阅读每台计算机的内容科学家应该知道浮点算术

When i print

(new BigDecimal(5) * new BigDecimal(0.049))​

It gives

0.24500000000000000943689570931383059360086917877197265625

When i round it using the ROUND_HALF_EVEN

(new BigDecimal(5) * new BigDecimal(0.049)).setScale(2, BigDecimal.ROUND_HALF_EVEN)​

It prints

0.25

So my confusion is, shouldn't it round to even number, so instead of 0.25 shouldn't it be 0.24. Please help with this confusion. Thanks!

解决方案

The REAL issue here is that you used the wrong constructor for the BigDecimal.

(new BigDecimal(5) * new BigDecimal("0.049")).setScale(2, BigDecimal.ROUND_HALF_EVEN)​

will do what you want.

The problem is that 0.049 is a floating point literal, and that value is not representable exactly as a floating point value (neither float nor double), introducing a miniscule error that in this case is meaningful.

By using the constructor that accepts a String argument you avoid the conversion through floating point and get the exact value you intended.

Floating point arithmetic on computers is fraught with nasty unexpected behaviors due to its limited precision. If you want to learn more about the pitfalls, read What Every Computer Scientist Should Know About Floating-Point Arithmetic

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