题
我如何计算A的平方根 Float
在 C#
, , 如同 Core.Sqrt
在XNA中?
解决方案
计算它 double
然后回到浮动。可能有点慢,但应该起作用。
(float)Math.Sqrt(inputFloat)
其他提示
讨厌这么说,但是0x5F3759DF似乎需要3倍,直到数学。我只是对计时器进行了一些测试。在循环访问预定阵列的前循环中的Math.sqrt约为80ms。 0x5F3759DF在相同情况下导致180+ms
使用释放模式优化进行了几次测试。
来源下:
/*
================
SquareRootFloat
================
*/
unsafe static void SquareRootFloat(ref float number, out float result)
{
long i;
float x, y;
const float f = 1.5F;
x = number * 0.5F;
y = number;
i = *(long*)&y;
i = 0x5f3759df - (i >> 1);
y = *(float*)&i;
y = y * (f - (x * y * y));
y = y * (f - (x * y * y));
result = number * y;
}
/*
================
SquareRootFloat
================
*/
unsafe static float SquareRootFloat(float number)
{
long i;
float x, y;
const float f = 1.5F;
x = number * 0.5F;
y = number;
i = *(long*)&y;
i = 0x5f3759df - (i >> 1);
y = *(float*)&i;
y = y * (f - (x * y * y));
y = y * (f - (x * y * y));
return number * y;
}
/// <summary>
/// The main entry point for the application.
/// </summary>
[STAThread]
static void Main()
{
int Cycles = 10000000;
Random rnd = new Random();
float[] Values = new float[Cycles];
for (int i = 0; i < Cycles; i++)
Values[i] = (float)(rnd.NextDouble() * 10000.0);
TimeSpan SqrtTime;
float[] Results = new float[Cycles];
DateTime Start = DateTime.Now;
for (int i = 0; i < Cycles; i++)
{
SquareRootFloat(ref Values[i], out Results[i]);
//Results[i] = (float)Math.Sqrt((float)Values[i]);
//Results[i] = SquareRootFloat(Values[i]);
}
DateTime End = DateTime.Now;
SqrtTime = End - Start;
Console.WriteLine("Sqrt was " + SqrtTime.TotalMilliseconds.ToString() + " long");
Console.ReadKey();
}
}
var result = Math.Sqrt((double)value);
private double operand1;
private void squareRoot_Click(object sender, EventArgs e)
{
operand1 = Math.Sqrt(operand1);
this.textBox1.Text = operand1.ToString();
}
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