您好,我不熟悉多线程并试图将其合并到我的程序中,以便让特定的计算更快地收敛。但是当我像下面这样创建两个线程时
Thread firstThread = new Thread(() => { p0 = f(changePoint, result); });
Thread secondThread = new Thread(() => { p1 = f(changePoint + 1, result); });
firstThread.Start();
secondThread.Start();
firstThread.Join();
secondThread.Join();
p0 和 p1 中的值相同(p0 与 p1 具有相同的值)。现在,如果我取消多线程并像这样调用它们:
p0 = f(changePoint, result);
p1 = f(changePoint + 1, result);
返回不同的值,一切正常。
我错过了什么?
f(x,y) 的代码
public double f(double x,double result)
{
double temp = PCAcont.Future2Yrs(x).Last().FirstOrDefault().StatNetWorthToAssets.GetValueOrDefault();
return temp - result;
}
主要方法
public double SecantMethod(double prec, int stepsCutoff, double changePoint, double result)
{
double p2, p1 = 0, p0 = 0;
int i;
Thread firstThread = new Thread(() => { p0 = f(changePoint, result); });
Thread secondThread = new Thread(() => { p1 = f(changePoint + 1, result); });
firstThread.Start();
secondThread.Start();
firstThread.Join();
secondThread.Join();
//p0 = f(changePoint, result);
//p1 = f(changePoint + 1, result);
p2 = p1 - f(p1, result) * (p1 - p0) / (f(p1, result) - f(p0, result));
for (i = 0; System.Math.Abs(p2 - p1) > prec && i < stepsCutoff; i++)
{
p0 = p1;
p1 = p2;
p2 = p1 - f(p1, result) * (p1 - p0) / (f(p1, result) - f(p0,result));
}
if (i < stepsCutoff)
return p2;
else
{
System.Diagnostics.Debug.WriteLine("{0}.The method did not converge", p2);
return double.NaN;
}
}