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确定javascript异步操作的结束(determining the end of asynchronous operations javascript)

 如果我有一个传递了这个函数的函数：  
function(work) {
   work(10);
   work(20);
   work(30);
}
 
 （可以有任意数量的work电话，其中包含任何数字。）  
 work性能一些异步活动 - 比如，对于这个例子，它只是一个timeout 。 我完全可以控制完成此操作的工作（事实上，它的定义一般）。  
 确定所有work通话何时完成的最佳方法是什么？  
 
 我的当前方法在调用工作时递增计数器，在完成时递减计数器，并在计数器为0时触发all work done事件（在每次递减后检查）。 但是，我担心这可能是某种竞争条件。 如果不是这样，请告诉我原因，这将是一个很好的答案。 

If I have a function that's passed this function: 
function(work) {
   work(10);
   work(20);
   work(30);
}
 
(There can be any number of work calls with any number in them.) 
work performance some asynchronous activity—say, for this example, it just is a timeout. I have full control over what work does on the completion of this operation (and, in fact, its definition in general). 
What's the best way of determining when all the calls to work are done? 
 
My current method increments a counter when work is called and decrements it when it completes, and fires the all work done event when the counter is 0 (this is checked after every decrement). However, I worry that this could be a race condition of some sort. If that is not the case, do show my why and that would be a great answer.

原文：https://stackoverflow.com/questions/6852059

更新时间：2022-04-04 07:04

最满意答案

 这是另一个版本，它带来了显着的改进：  
   function cohens_kappa2(x::Vector{Int}, k::Int)
     d = Dict{Int,Int}()
     n = length(x)
     c1 = Int[]
     pnew = 0
     for i=1:n
       p = get(d,x[i],0)
       if p>0
         c1[p] += 1
       else
         pnew += 1
         d[x[i]] = pnew
         push!(c1,1)
       end
     end
     c2 = zeros(Int,pnew)
     for i=(k+1):n
       if x[i-k]==x[i] c2[d[x[i]]] += 1 ; end
     end
     num, dentmp = 0.0, 0.0
     for i=1:pnew
       pjjk = c2[i]/(n-k)
       pj = c1[i] / n
       num += pjjk - pj^2
       dentmp += pj^2
     end
     return (num / (1.0-dentmp))
   end
 
 一般来说，几乎总是可以进行优化，但就像从自然界中提取石油一样，它会增加程序员的成本和工作量。  
 在一个测试用例中，上面给了我5倍到10倍的加速。 您的数据结果如何？ 

Here is another version, which gives a significant improvement: 
   function cohens_kappa2(x::Vector{Int}, k::Int)
     d = Dict{Int,Int}()
     n = length(x)
     c1 = Int[]
     pnew = 0
     for i=1:n
       p = get(d,x[i],0)
       if p>0
         c1[p] += 1
       else
         pnew += 1
         d[x[i]] = pnew
         push!(c1,1)
       end
     end
     c2 = zeros(Int,pnew)
     for i=(k+1):n
       if x[i-k]==x[i] c2[d[x[i]]] += 1 ; end
     end
     num, dentmp = 0.0, 0.0
     for i=1:pnew
       pjjk = c2[i]/(n-k)
       pj = c1[i] / n
       num += pjjk - pj^2
       dentmp += pj^2
     end
     return (num / (1.0-dentmp))
   end
 
In general, optimization is almost always possible, but like extracting oil from nature, it comes with growing costs and effort for the programmer. 
On a test case the above gave me 5x to 10x speedup. What is the result for your data?

确定javascript异步操作的结束(determining the end of asynchronous operations javascript)

最满意答案

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