围捕在C＃2位小数 [英] Rounding up to 2 decimal places in C#

问题描述

我有一个十进制数可以是这样的：

I have a decimal number which can be like the following:

189.182

我想这一轮的最多的2位小数，所以输出将是以下内容：

I want to round this up to 2 decimal places, so the output would be the following:

189.19

时有内置的功能，这在数学类，或者其他什么东西？我知道天花板函数存在，但这似乎并没有做我想做的 - 它会舍入到最近INT，所以就在这种情况下，189

Is there built in functionality for this in the Math class, or something else? I know the ceiling function exists but this doesn't seem to do what I want - it'll round to the nearest int, so just '189' in this case.

推荐答案

乘以100，呼叫天花板，除以100做什么，我想你问的

Multiply by 100, call ceiling, divide by 100 does what I think you are asking for

public static double RoundUp(double input, int places)
{
    double multiplier = Math.Pow(10, Convert.ToDouble(places));
    return Math.Ceiling(input * multiplier) / multiplier;
}

使用看起来像：

RoundUp(189.182, 2);

这的工作原理是将小数点右边2位（所以它是对过去8的右侧）然后执行所述天花板的操作，然后将小数点回到其原来的位置。

This works by shifting the decimal point right 2 places (so it is to the right of the last 8) then performing the ceiling operation, then shifting the decimal point back to its original position.

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