为什么Convert.ToInt32（1.0 / 0.00004）！=（Int32）在（1.0 / 0.00004） [英] Why Convert.ToInt32(1.0/0.00004) != (Int32)(1.0/0.00004)

问题描述

为什么这个code http://ideone.com/YRcICG

Why this code http://ideone.com/YRcICG

void Main()
{   
    double a = 0.00004;
    Int32 castToInt = (Int32)(1.0/a);
    Int32 convertToInt = Convert.ToInt32(1.0/a);

    Console.WriteLine("{0} {1:F9} {2:F9}", castToInt == convertToInt, castToInt, convertToInt);

    Console.WriteLine((((int)(1.0/(1.0/25000))) == 24999));
}

结果

假24999,000000000 25000,000000000
  真

False 24999,000000000 25000,000000000
True

在CLR上下文/ C＃实现

in context of CLR/C# implementation

推荐答案

诀窍在于方式的双重重新presented左右（1.0 / A）将被重新psented通过以下方式$ P $

The trick lies in the way the double is represented so (1.0/a) will be represented in the following way:

（1.0 / A）= 24999.99999999999636202119290828704833984375。

(1.0/a) = 24999.99999999999636202119290828704833984375.

当你使用CAST你得到这个数字的只有前半部分，而转换方法的工作方式不一样，这里是code为转换方法：

When you use cast you get only the first part of this number, while the convert Method works in a different way, here is the code for the Convert method:

public static int ToInt32(double value)
{
    if (value >= 0.0)
    {
        if (value < 2147483647.5)
        {
            int num = (int)value;
            double num2 = value - (double)num;
            if (num2 > 0.5 || (num2 == 0.5 && (num & 1) != 0))
            {
                num++;
            }
            return num;
        }
    }
    else
    {
        if (value >= -2147483648.5)
        {
            int num3 = (int)value;
            double num4 = value - (double)num3;
            if (num4 < -0.5 || (num4 == -0.5 && (num3 & 1) != 0))
            {
                num3--;
            }
            return num3;
        }
    }
    throw new OverflowException(Environment.GetResourceString("Overflow_Int32"));
}

正如你可以看到有一个if语句来检查铸造的价值和原双之间的差异，在您的例子是：

As you can see there is an if statement that checks the difference between casted value and original double, in your example it is:

int num = (int)value;
double num2 = value - (double)num;

24999.99999999999636202119290828704833984375 - 24999 > 0.5

，所以你得到的增量。

, hence you get the increment.

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