iOS 轮一个浮点数
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iOS round a float
提问by ghiboz
I have a floating point number that have more decimal digits, for example:
我有一个具有更多十进制数字的浮点数，例如：
float fRes = 10.0 / 3.0;
actually the fRes value is 3.3333333333333 it's possible set for example 2 decimal digits:
实际上 fRes 值是 3.3333333333333 可以设置例如 2 个十进制数字：
float fRes = 10.0 / 3.0;
// fRes is 3.333333333333333333333333
float fResOk = FuncRound( fRes, 2 );
// fResOk is 3.33
thanks in advance
提前致谢
回答by Nathan Day
I don't know where you are using this rounded number, but you should only round your value when displaying it to the user, there are C based format string ways to round floating point numbers for example
我不知道你在哪里使用这个四舍五入的数字，但你应该只在向用户显示时四舍五入你的值，例如有基于 C 的格式字符串方法来四舍五入浮点数
[NSString stringWithFormat:@"%.2f", value];
as you may have already read, floating point number are approximations of real numbers, so doing fResOk = roundf( fRes*100.0)/100.0;may not give you 3.33 but a number which is just as close as you can get with floating point number to 3.33.
您可能已经读过，浮点数是实数的近似值，因此这样做fResOk = roundf( fRes*100.0)/100.0;可能不会给您 3.33，而是一个与浮点数与 3.33 最接近的数字。
回答by gaige
Assuming that you're looking for the correct function to round to a certain number of digits, you'll probably find it easiest to do the following:
假设您正在寻找四舍五入到特定位数的正确函数，您可能会发现执行以下操作最简单：
fResOk = roundf( fRes*100.0)/100.0;
That will multiply the value by 100(giving you your 2 digits of significance), round the value, and then reduce it back to the magnitude you originally started with.
这会将值乘以100（为您提供 2 位重要数字），舍入该值，然后将其减少到您最初开始时的大小。
