C++ 中的 Round() 在哪里?
声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow
原文地址: http://stackoverflow.com/questions/554204/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Where is Round() in C++?
提问by Jason Punyon
Duplicate of: round() for float in C++
重复:round() 用于 C++ 中的浮点数
I'm using VS2008 and I've included math.h but I still can't find a round function. Does it exist?
我正在使用 VS2008 并且我已经包含了 math.h 但我仍然找不到圆形函数。它存在吗?
I'm seeing a bunch of "add 0.5 and cast to int" solutions on google. Is that the best practice?
我在谷歌上看到了一堆“添加 0.5 并转换为 int”的解决方案。这是最好的做法吗?
回答by Patrick Glandien
You may use C++11's std::round()
.
您可以使用 C++11 的std::round()
.
If you are still stuck with older standards, you may use std::floor()
, which always rounds to the lower number, and std::ceil()
, which always rounds to the higher number.
如果您仍然坚持使用旧标准,您可以使用std::floor()
,它总是四舍五入到较低的数字,而std::ceil()
,它总是四舍五入到较高的数字。
To get the normal rounding behaviour, you would indeed use floor(i + 0.5)
.
要获得正常的舍入行为,您确实会使用floor(i + 0.5)
.
This way will give you problems with negative numbers, a workaround for that problem is by using ceil() for negative numbers:
这种方式会给您带来负数问题,该问题的解决方法是对负数使用 ceil() :
double round(double number)
{
return number < 0.0 ? ceil(number - 0.5) : floor(number + 0.5);
}
Another, cleaner, but more resource-intensive, way is to make use of a stringstream and the input-/output-manipulators:
另一种更简洁但更占用资源的方法是使用字符串流和输入/输出操作符:
#include <iostream>
#include <sstream>
double round(double val, int precision)
{
std::stringstream s;
s << std::setprecision(precision) << std::setiosflags(std::ios_base::fixed) << val;
s >> val;
return val;
}
Only use the second approach if you are not low on resources and/or need to have control over the precision.
如果您不缺乏资源和/或需要控制精度,请仅使用第二种方法。
回答by Bill the Lizard
Using floor(num + 0.5)
won't work for negative numbers. In that case you need to use ceil(num - 0.5)
.
使用floor(num + 0.5)
不适用于负数。在这种情况下,您需要使用ceil(num - 0.5)
.
double roundToNearest(double num) {
return (num > 0.0) ? floor(num + 0.5) : ceil(num - 0.5);
}
回答by Sani Singh Huttunen
There actually isn't a round function in Microsoft math.h.
However you could use the static method Math::Round() instead.
(Depending on your project type.)
Microsoft math.h 中实际上没有舍入函数。
但是,您可以改用静态方法 Math::Round()。
(取决于您的项目类型。)
回答by Elroy
I don't know if it's best practice or not but using the 0.5 technique with a floor() seems to be the way to go.
我不知道这是否是最佳实践,但使用 0.5 技术和 floor() 似乎是要走的路。