このplinqのバグですか？

Question

PLINQ出力が順次処理とパラレルで異なる理由：Loop

私は10,000,000個の平方根の合計を追加したいです。ここでは3例のコードです：

ループの順次：

double sum = 0.0;
for(int i = 1;i<10000001;i++)
sum += Math.Sqrt(i);

.

出力は次のとおりです。 21081852648.717

現在パラレルを使用しています。ループ：

object locker = new object();
double total ;

Parallel.For(1,10000001,
()=>0.0,
(i,state,local)=> local+Math.Sqrt(i),
(local)=>
{
  lock(locker){ total += local; }
}
);

.

出力の出力は次のとおりです。 21081852648.7199

今PLINQを使用している

double tot =  ParallelEnumerable.Range(1, 10000000)
                .Sum(i => Math.Sqrt(i)); 

.

出力の出力は次のとおりです。 21081852648.72

PLINQ出力と並列の間に違いがあるのはなぜループの順次ですか？

Answer 1

I strongly suspect it's because arithmetic with doubles isn't truly associative. Information is potentially lost while summing values, and exactly what information is lost will depend on the order of the operations.

Here's an example showing that effect:

using System;

class Test
{
    static void Main()
    {
        double d1 = 0d;
        for (int i = 0; i < 10000; i++)
        {
            d1 += 0.00000000000000001;
        }
        d1 += 1;
        Console.WriteLine(d1);

        double d2 = 1d;
        for (int i = 0; i < 10000; i++)
        {
            d2 += 0.00000000000000001;
        }
        Console.WriteLine(d2);
    }
}

In the first case, we can add very small numbers lots of times until they become big enough to still be relevant when added to 1.

In the second case, adding 0.00000000000000001 to 1 always just results in 1 as there isn't enough information in a double to represent 1.00000000000000001 - so the final result is still just 1.

EDIT: I've thought of another aspect which could be confusing things. For local variables, the JIT compiler is able to (and allowed to) use the 80-bit FP registers, which means arithmetic can be performed with less information loss. That's not the case for instance variables which definitely have to be 64-bit. In your Parallel.For case, the total variable will actually be an instance variable in a generated class because it's captured by a lambda expression. This could change the results - but it may well depend on computer architecture, CLR version etc.