舍入十进制值 [英] Rounding decimal value

问题描述

使用 Math.Round（a，2）;

<当我用1.2小数点后两位数舍入时，结果是1.27。
当我做同样的1.375时，结果是1.38。

为什么不是四舍五入的1.275到1.28？

谢谢

解决方案

我不能重现您的问题：

  Math.Round（1.275m，2）=> 1.28m 
 Math.Round（1.375m，2）=> 1.38m 
  

我怀疑你声称你使用 decimal 值为false，而您使用 double 值来代替。 double 不能完全表示很多十进制值，所以当你写 1.275 时，实际上是1.27499 ... 1.375 是少数可表示的元素之一，所以它实际上是 1.375 。

如果您的代码关心精确的十进制表示，例如在处理金钱时，您必须使用 decimal ，而不是二进制浮点数如 double 或 float 。

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但是即使使用十进制表示法，对于许多用户来说，四舍五入也会出乎意料：
$ b $ pre $ Math.Round 1.265m，2）=> 1.26m
Math.Round（1.275m，2）=> 1.28m


默认 Math.Round 使用 MidpointRounding.ToEven ，也称为银行家轮。这样可以避免在 .5 之间累积偏差。

您可以使用 Round ，它采用舍入模式，并将其设置为 AwayFromZero 以获得您期望的行为

  Math.Round（1.275m，2，MidpointRounding.AwayFromZero）=> 1.28m 
  

I am facing problem while rounding the decimal value in C# using Math.Round(a, 2);

When I'm rounding 1.275 by 2 decimal points, the result is 1.27. When I'm doing the same for 1.375, the result is 1.38.

Why is it not rounding 1.275 to 1.28?

Thanks

解决方案

I cannot reproduce your problem:

Math.Round(1.275m, 2) => 1.28m
Math.Round(1.375m, 2) => 1.38m

I suspect that your claim that you use a decimal value is false, and that you use double value instead. double can't represent many decimal values exactly, so when you write 1.275, it's actually 1.27499... 1.375 is one of the few representable onces, so it's actually 1.375.

If your code cares about exact decimal representation, for example when you work on money, you must use decimal and not binary floating point such as double or float.

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But even if you use decimal representation, rounding behaves unexpectedly for many users:

Math.Round(1.265m, 2) => 1.26m
Math.Round(1.275m, 2) => 1.28m

By default Math.Round uses MidpointRounding.ToEven, also known as Banker's round. This avoids accumulating a bias from always rounding up at .5.

You can use an overload of Round that takes a rounding mode, and set it to AwayFromZero to get the behaviour you expect.

Math.Round(1.275m, 2, MidpointRounding.AwayFromZero) => 1.28m

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