Concurrency X10并行处理共享变量
如果我的问题不专业,请原谅。我正在阅读IBM x10的教程。下面是计算PI但让我困惑的代码:Concurrency X10并行处理共享变量,concurrency,parallel-processing,x10-language,Concurrency,Parallel Processing,X10 Language,如果我的问题不专业,请原谅。我正在阅读IBM x10的教程。下面是计算PI但让我困惑的代码: public static def countPoints(n: Int, rand: ()=>Double) { var inCircle: Double = 0.0; for (var j:Long = 1; j<=n; j++) { val x = rand(); val y = rand(); if (x*x +y*y <= 1.0) inCircl
public static def countPoints(n: Int, rand: ()=>Double) {
var inCircle: Double = 0.0;
for (var j:Long = 1; j<=n; j++) {
val x = rand();
val y = rand();
if (x*x +y*y <= 1.0) inCircle++;
}
return inCircle;
}
val N = args.size() > 0 ? Long.parse(args(0)) : 100000;
val THREADS = args.size() > 1 ? Int.parse(args(1)) : 4;
val nPerThread = N/THREADS;
val inCircle = new Array[Long](1..THREADS);
finish for(var k: Int =1; k<=THREADS; k++) {
val r = new Random(k*k + k + 1);
val rand = () => r.nextDouble();
val kk = k;
async inCircle(kk) = countPoints(nPerThread,rand);
}
var totalInCircle: Long = 0;
for(var k: Int =1; k<=THREADS; k++) {
totalInCircle += inCircle(k);
}
val pi = (4.0*totalInCircle)/N;
publicstaticdef countPoints(n:Int,rand:()=>Double){
内圆变量:双=0.0;
对于(varj:Long=1;j1?Int.parse(args(1)):4;
val nPerThread=N个线程;
val inCircle=新数组[Long](1..THREADS);
完成(变量k:Int=1;kr.nextDouble();
val kk=k;
异步内循环(kk)=计数点(nPerThread,rand);
}
var totalInCircle:长=0;
对于(var k:Int=1;k很好,您在这里担心可能的竞争条件。在并行调用随机数生成器时,它经常被忽略
幸运的是,这个例子没有RNG竞态条件。for循环的每次迭代都会创建一个随机数生成器的新实例(并为其种子),并生成一个线程。因为countPoints
调用它自己的RNG,所以这里没有竞态条件