如何对同一阵列中的随机位置执行多线程操作？

我有一个进程在随机位置上运行大量任务,Array并希望通过使用多线程来加快速度.

它本质上做的是随机化"阵列"中的位置,检查其近似环境的值,并在满足一些特定条件时改变随机位置值.

是否有可能运行像

Parallel.For(0, n, s => { });

循环而不是下面显示的while代码块来优化这个函数,一个代码块怎么样呢？

我一直在考虑为所选元素使用一些"忙"属性,但这实际上使得问题可能需要更加复杂.

public void doStuffTothisArray(ref int[,,] Array, ref IGenerator randomGenerator, int loops)
{
    int cc = 0;
    int sw = 0;
    do
    {
        if (doStuffOnRandomPositions(ref Array, ref randomGenerator))
            sw++; //if stuff was made counter

        if ((cc % (loops / 10)) == 0)
            Console.Write("{0} % \t", (cc / (loops / 10)) * 10); //some loading info

        cc++; //count iterations
    } while (cc < loops);
    Console.WriteLine("Stuff altered in {0} iterations: {1}", loops, sw);
}

发布编辑:

划分阵列并分配工作会破坏阵列的动态,因为它需要是一个完整的系统.

这是dostuff的原型..()

public static bool doStuffOnRandomPositions(ref lattice A, ref IGenerator rr)
{
    position firstPos = new position(rr.Next(0, A.n_size),rr.Next(0, A.n_size),rr.Next(0, A.n_size));
    position secondPos = randomNeighbour(ref A, firstPos, ref rr);

    //checks the closest 3d neighbours indexer the lattice
    //Console.WriteLine("first:[{0},{1},{2}]\nsecond:[{3},{4},{5}]\n", firstPos.x, firstPos.y, firstPos.z, secondPos.x, secondPos.y, secondPos.z);

    //  get values at coordinates
    bool first = A.latticeArray[firstPos.x, firstPos.y, firstPos.z];
    bool second = A.latticeArray[secondPos.x,secondPos.y,secondPos.z];

    if (first == second) //don't bother if they are equal states
        return false;

    //  checks the energies in surroundings for an eventual spin switch
    int surrBefore = surroundCheck(ref A, firstPos, first) ; // - surroundCheck(ref A, secondPos, second));
    int surrAfter = surroundCheck(ref A, firstPos, !first) ; // - surroundCheck(ref A, secondPos, !second));

    if (surrAfter < surrBefore) //switch spin states if lower total energy
    {
        A.latticeArray[firstPos.x, firstPos.y, firstPos.z] = !first;
        A.latticeArray[secondPos.x, secondPos.y, secondPos.z] = !second;
        return true;
    }
    else if ((surrAfter == surrBefore) & latticeDistribution(ref rr))   //TEMPORARY
    {
        A.latticeArray[firstPos.x, firstPos.y, firstPos.z] = !first;        //TEMPORARY
        A.latticeArray[secondPos.x, secondPos.y, secondPos.z] = !second;    //TEMPORARY
        return true;
    }
    else
        return false;
} //FIX SWITCH PROBABILITIES

在这里,晶格类应该表示包含其属性的"数组".由于我对c#方法的经验不足,示例解决方案代码非常感谢.