Math.Round有点烦人！ [英] Math.Round is being a bit annoying!

问题描述

所以我读过如何围绕数字的文章。我基本上为控制台打印出了一点钱。这是总和：

So I''ve read the article in how to round numbers. And I''ve basically made a little sum for the console to print out. This is the sum:

int rndx = (int)Math.Round((decimal)(53 / 32), 0, MidpointRounding.AwayFromZero);
Console.WriteLine(rndx.ToString());

每次运行此代码时，它都会返回'' ''。虽然它应该是两个。



我甚至检查了我的计算器，其中53/32 = 1,65625。这将被四舍五入为2，因为它超过数字''x.5''。



任何答案？

Everytime I run this code, it returns as ''one''. While it should be two.

I''ve even checked my calculator which says that 53 / 32 = 1,65625. Which would be rounded up to 2 since it''s over number ''x.5''.

Any answers?

推荐答案

我希望它能返回一个！ ：笑：



为什么？简单：



53和32是整数。所以53/32也是一个整数，值为1.



整数值的 double 强制转换不会恢复丢失的分数数据...

尝试：

I expect it to return one! :laugh:

Why? Simples:

53 and 32 are integers. so 53 / 32 is also an integer, value 1.

The double cast of an integer value will not restore lost fractional data...
Try:

int rndx = (int)Math.Round(((decimal)53 / (decimal)32), 0, MidpointRounding.AwayFromZero);

或

Or

int rndx = (int)Math.Round((53M / 32M), 0, MidpointRounding.AwayFromZero);

其中指定十进制常数。



手指输入 - 脑部疾病。 double表示十进制：doh： - OriginalGriff [/ edit]

Which specifies decimal constants.

[edit]Fingers-typing-ahead-of-brain disease. "double" for "decimal" :doh: - OriginalGriff[/edit]

Math.Round [ ^ ]返回十进制值，而不是整数！点击链接！



参见其他方法：

Math.Floor（） [ ^ ]

Math.Ceiling（） [ ^ ]

Math.Round[^] returns decimal value, not integer! Follow the link!

See other methods:
Math.Floor()[^]
Math.Ceiling()[^]

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