矩阵乘法？
我在这里尝试的是两个 500x500 双精度矩阵的乘法。 我收到空引用异常！
你能看一下吗
void Topt(double[][] A, double[][] B, double[][] C) {
var source = Enumerable.Range(0, N);
var pquery = from num in source.AsParallel()
select num;
pquery.ForAll((e) => Popt(A, B, C, e));
}
void Popt(double[][] A, double[][] B, double[][] C, int i) {
double[] iRowA = A[i];
double[] iRowC = C[i];
for (int k = 0; k < N; k++) {
double[] kRowB = B[k];
double ikA = iRowA[k];
for (int j = 0; j < N; j++) {
iRowC[j] += ikA * kRowB[j];
}
}
}
提前致谢
What i am trying here is multiplication of two 500x500 matrices of doubles.
And i am getting null reference exception !
Could u take a look into it
void Topt(double[][] A, double[][] B, double[][] C) {
var source = Enumerable.Range(0, N);
var pquery = from num in source.AsParallel()
select num;
pquery.ForAll((e) => Popt(A, B, C, e));
}
void Popt(double[][] A, double[][] B, double[][] C, int i) {
double[] iRowA = A[i];
double[] iRowC = C[i];
for (int k = 0; k < N; k++) {
double[] kRowB = B[k];
double ikA = iRowA[k];
for (int j = 0; j < N; j++) {
iRowC[j] += ikA * kRowB[j];
}
}
}
Thanks in advance
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评论（1）
既然你的空指针问题已经解决了，为什么不提供一个小的性能提示;)你可以尝试的一件事是缓存不经意的算法。对于 2k x 2k 矩阵，我得到了 24.7 秒的递归缓存不经意变量和 50.26 秒，在 C 中的 e8400 @3ghz 单线程上使用简单的迭代方法（两者都可以明显地进一步优化，为编译器提供一些更好的参数并确保 SSE被使用等）。虽然 500x500 相当小，所以您必须尝试一下这是否能给您带来明显的改进。
不过，递归通常更容易实现多线程。
哦，但自从你用 c# 编写以来，最重要的事情是：阅读 this - 它适用于 java，但同样适用于任何现代 JIT。
Since your nullpointer problem was already solved, why not a small performance tip ;) One thing you could try would be a cache oblivious algorithm. For 2k x 2k matrices I get 24.7sec for the recursive cache oblivious variant and 50.26s with the trivial iterative method on a e8400 @3ghz single threaded in C (both could be further optimized obviously, some better arguments for the compiler and making sure SSE is used etc.). Though 500x500 is quite small so you'd have to try if that gives you noticeable improvements.
The recursive one is easier to make multithreaded usually though.
Oh but the most important thing since you're writing in c#: Read this - it's for java, but the same should apply for any modern JIT..