C++：如何将双精度舍入为整数？
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/9695329/
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
C++: How to round a double to an int?
提问by midnightBlue
I have a double (call it x), meant to be 55 but in actuality stored as 54.999999999999943157 which I just realised.
我有一个double（称为x），本来是55，但实际上存储为54.999999999999943157，我刚刚意识到。
So when I do
所以当我做
double x = 54.999999999999943157;
int y = (int) x;
y = 54 instead of 55!
y = 54 而不是 55！
This puzzled me for a long time. How do I get it to correctly round?
这让我困惑了很长时间。我如何让它正确舍入？
回答by Moritz
add 0.5 before casting (if x > 0) or subtract 0.5 (if x < 0), because the compiler will always truncate.
在强制转换之前加 0.5（如果 x > 0）或减去 0.5（如果 x < 0），因为编译器总是会截断。
float x = 55; // stored as 54.999999...
x = x + 0.5 - (x<0); // x is now 55.499999...
int y = (int)x; // truncated to 55
C++11 also introduces std::round, which likely uses a similar logic of adding 0.5 to |x| under the hood (see the link if interested) but is obviously more robust.
C++11 还引入了std::round，它可能使用类似的逻辑将 0.5 添加到 |x| 在引擎盖下（如果有兴趣，请参阅链接）但显然更强大。
A follow up question might be whythe float isn't stored as exactly 55. For an explanation, see thisstackoverflow answer.
一个后续问题可能是为什么浮点数没有准确存储为 55。有关解释，请参阅此stackoverflow 答案。
回答by MK.
Casting is not a mathematical operation and doesn't behave as such. Try
铸造不是数学运算，也不是这样的。尝试
int y = (int)round(x);
回答by Pubby
Casting to an inttruncates the value. Adding 0.5causes it to do proper rounding.
转换为 an 会int截断该值。添加0.5会导致它进行适当的舍入。
int y = (int)(x + 0.5);
回答by Pubby
It is worth noting that what you're doing isn't rounding, it's casting. Casting using (int) xtruncates the decimal value of x. As in your example, if x = 3.9995, the .9995gets truncated and x = 3.
值得注意的是，您所做的不是四舍五入，而是铸造。使用(int) x强制转换会截断 的十进制值x。在您的示例中，如果x = 3.9995，.9995则被截断，而x = 3.
As proposed by many others, one solution is to add 0.5to x, and then cast.
正如许多其他人提出的那样，一种解决方案是添加0.5到x，然后进行强制转换。
回答by zik
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
double x=54.999999999999943157;
int y=ceil(x);//The ceil() function returns the smallest integer no less than x
return 0;
}
