并行ForEach在不同的Ocasions中给出不同的结果

Question

我从来没有用过Parallel.ForEach但我玩过它并发现了这种情况。

我运行并行循环与枚举范围首先(0,1)然后(0,1,2)然后我再次运行第二个是睡着了线程后（代码MSDN上https://msdn.microsoft.com/en-us/library/dd997393(v=vs.110).aspx发现并与subtotal *=2编辑试图理解它在做什么）为200毫秒，并且将结果然后不同

如果Thread.sleep(200)没有被注释掉这是结果

result 1 = 2 
result 2 = 6 
result 3 = 4 

如果Thread.sleep(200)被注释掉是这种结果

result 1 = 2 
result 2 = 6 
result 3 = 6 

下面是代码

Stopwatch timer = new Stopwatch(); 
int[] nums = Enumerable.Range(0, 1).ToArray(); 
long total = 0; 
for (int i = 0; i < 2; i++) 
{ 
    timer.Restart(); 
    total = 0; 
    if (i == 0) nums = Enumerable.Range(0, 1).ToArray(); 
    if (i == 1) nums = Enumerable.Range(0, 2).ToArray(); 
    Parallel.ForEach<int, long>(nums,() => 0,(j, loop, subtotal) => 
    { 
     subtotal += 1; 
     subtotal *= 2; 
     return subtotal; 
    },(finalResult) => Interlocked.Add(ref total, finalResult)); 

    Console.WriteLine("The total from Parallel.ForEach is {0:N0} and took {1}", total, timer.Elapsed); 
    timer.Stop(); 
    //Thread.Sleep(200); 
} 

timer.Restart(); 
nums = Enumerable.Range(0, 2).ToArray(); 
total = 0; 
Parallel.ForEach<int, long>(nums,() => 0, (j, loop, subtotal) => 
{ 
    subtotal += 1; 
    subtotal *= 2; 
    return subtotal; 
},(finalResult) => Interlocked.Add(ref total, finalResult)); 

Console.WriteLine("The total from Parallel.ForEach is {0:N0} and took {1}", total, timer.Elapsed); 
timer.Stop(); 

我想它已经得到了与工作在彼此线程做，但这似乎是一个错误

注意我也看看Simulation gives different result with normal for loop Vs Parallel For
这是怎么发生的？

Answer 1

因为这个代码是不明确的：

Parallel.ForEach<int, long>(nums,() => 0,(j, loop, subtotal) => 
{ 
    subtotal += 1; 
    subtotal *= 2; 
    return subtotal; 
},(finalResult) => Interlocked.Add(ref total, finalResult)); 

在这方面，如果单线程执行两个迭代，那么你得到的结果6.实际上，你这样做：

subTotal = 0; //From init 
subTotal += 1; //=1 First iteration 
subTotal *= 2; //=2 First iteration 
subTotal += 1; //=3 Second iteration 
subTotal *= 2; //=6 Second iteration 
total += subTotal; //=6 End gathering (actually interlocked) 

但是，如果两个线程共享工作，你会得到

subTotal1 = 0; //From init 
subTotal2 = 0; //From init 
subTotal2 += 1; //=1 
subTotal1 += 1; //=1 
subTotal1 *= 2; //=2 
subTotal2 *= 2; //=2 
total += subTotal1 //=2 End gathering 1 (interlocked) 
total += subTotal2 //=4 End gathering 2 (interlocked)