iOS围绕一个浮动 [英] iOS round a float
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问题描述
float fRes = 10.0 / 3.0;我有一个浮点数有更多的小数位,例如:
实际上fRes的值是3.3333333333333
它可能被设置为例子2的十进制数字: p>
float fRes = 10.0 / 3.0;
// fRes是3.333333333333333333333333
float fResOk = FuncRound(fRes,2);
// fResOk是3.33
提前致谢
<假设你正在寻找正确的函数来四舍五入到一定数量的数字,你可能会发现最容易做到以下几点:
fResOk = roundf(fRes * 100.0)/100.0;
这将乘以 100
(给你你的2位数的重要性),四舍五入的价值,然后减少它回到您最初开始的数量级。
I have a floating point number that have more decimal digits, for example:
float fRes = 10.0 / 3.0;
actually the fRes value is 3.3333333333333 it's possible set for example 2 decimal digits:
float fRes = 10.0 / 3.0;
// fRes is 3.333333333333333333333333
float fResOk = FuncRound( fRes, 2 );
// fResOk is 3.33
thanks in advance
解决方案
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:
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.
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