C# 如何将 double 转换为最接近的整数值？
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How might I convert a double to the nearest integer value?
提问by leora
How do you convert a double into the nearest int?
你如何将双精度转换为最近的整数？
采纳答案by nickf
Use Math.round(), possibly in conjunction with MidpointRounding.AwayFromZero
使用Math.round()，可能与MidpointRounding.AwayFromZero
eg:
例如：
Math.Round(1.2) ==> 1
Math.Round(1.5) ==> 2
Math.Round(2.5) ==> 2
Math.Round(2.5, MidpointRounding.AwayFromZero) ==> 3
回答by John Sheehan
double d = 1.234;
int i = Convert.ToInt32(d);
Handles rounding like so:
像这样处理舍入：
rounded to the nearest 32-bit signed integer. If value is halfway between two whole numbers, the even number is returned; that is, 4.5 is converted to 4, and 5.5 is converted to 6.
四舍五入到最接近的 32 位有符号整数。如果 value 介于两个整数之间，则返回偶数；即4.5转4，5.5转6。
回答by John Sheehan
double d;
int rounded = (int)Math.Round(d);
回答by Mark Jones
I'm developing a scientific calculator that sports an Int button. I've found the following is a simple, reliable solution:
我正在开发一个带有 Int 按钮的科学计算器。我发现以下是一个简单可靠的解决方案：
double dblInteger;
if( dblNumber < 0 )
dblInteger = Math.Ceiling(dblNumber);
else
dblInteger = Math.Floor(dblNumber);
Math.Round sometimes produces unexpected or undesirable results, and explicit conversion to integer (via cast or Convert.ToInt...) often produces wrong values for higher-precision numbers. The above method seems to always work.
Math.Round 有时会产生意外或不受欢迎的结果，并且显式转换为整数（通过强制转换或 Convert.ToInt...）通常会为更高精度的数字产生错误的值。上述方法似乎总是有效。
回答by szymcio
You can also use function:
您还可以使用功能：
//Works with negative numbers now
static int MyRound(double d) {
if (d < 0) {
return (int)(d - 0.5);
}
return (int)(d + 0.5);
}
Depending on the architecture it is several times faster.
根据架构，它要快几倍。
回答by Trent
I know this question is old, but I came across it in my search for the answer to my similar question. I thought I would share the very useful tip that I have been given.
我知道这个问题很老，但我在寻找类似问题的答案时遇到了它。我想我会分享我得到的非常有用的提示。
When converting to int, simply add .5to your value before downcasting. As downcasting to intalways drops to the lower number (e.g. (int)1.7 == 1), if your number is .5or higher, adding .5will bring it up into the next number and your downcast to intshould return the correct value. (e.g. (int)(1.8 + .5) == 2)
转换为 int 时，只需.5在向下转换之前添加到您的值。由于向下转换到int总是下降到较低的数字（例如(int)1.7 == 1），如果您的数字是.5或更高，则添加.5会将其带入下一个数字，并且向下转换到int应该返回正确的值。（例如(int)(1.8 + .5) == 2）
回答by zwcloud
For Unity, use Mathf.RoundToInt.
对于 Unity，请使用Mathf.RoundToInt。
using UnityEngine;
public class ExampleScript : MonoBehaviour
{
void Start()
{
// Prints 10
Debug.Log(Mathf.RoundToInt(10.0f));
// Prints 10
Debug.Log(Mathf.RoundToInt(10.2f));
// Prints 11
Debug.Log(Mathf.RoundToInt(10.7f));
// Prints 10
Debug.Log(Mathf.RoundToInt(10.5f));
// Prints 12
Debug.Log(Mathf.RoundToInt(11.5f));
// Prints -10
Debug.Log(Mathf.RoundToInt(-10.0f));
// Prints -10
Debug.Log(Mathf.RoundToInt(-10.2f));
// Prints -11
Debug.Log(Mathf.RoundToInt(-10.7f));
// Prints -10
Debug.Log(Mathf.RoundToInt(-10.5f));
// Prints -12
Debug.Log(Mathf.RoundToInt(-11.5f));
}
}
public static int RoundToInt(float f) { return (int)Math.Round(f); }
回答by Gru
Methods in other answers throw OverflowExceptionif the float value is outside the Int range. https://docs.microsoft.com/en-us/dotnet/api/system.convert.toint32?view=netframework-4.8#System_Convert_ToInt32_System_Single_
OverflowException如果浮点值超出 Int 范围，则其他答案中的方法将抛出。https://docs.microsoft.com/en-us/dotnet/api/system.convert.toint32?view=netframework-4.8#System_Convert_ToInt32_System_Single_
int result = 0;
try {
result = Convert.ToInt32(value);
}
catch (OverflowException) {
if (value > 0) result = int.MaxValue;
else result = int.Minvalue;
}
