将int除以10舍入为最接近的int(Rounding up an int divided by 10 to nearest int)

将int除以10舍入为最接近的int(Rounding up an int divided by 10 to nearest int)

我想将int除以10得到一个最接近舍入的整数，所以12应该给1，17应该给2

继承我的代码：

int BonusValue; int Str; BonusValue = Convert.ToInt32(Math.Round(Convert.ToDouble(Str) / 10));

这会有用吗？

I want to divide an int by 10 and get a whole number closest to rounded one, so 12 should give 1 and 17 should give 2

Heres my code:

int BonusValue; int Str; BonusValue = Convert.ToInt32(Math.Round(Convert.ToDouble(Str) / 10));

Will this work?

最满意答案

是的，那应该有用。

但是，如果将整数中的一半舍入步骤（在您的情况下为5）添加，则会更容易。 然后你可以做简单的整数除法：

BonusValue = (Str+5) / 10

Yes, that should work.

It is however much easier if you add half the rounding step, in your case 5, to the integer. Then you can do simple integer division:

BonusValue = (Str+5) / 10