Parallel.For: Example

Here's a simple example I've devised to illustrate Parallel.For and to compare it with the 'normal' sequential For statement.

On my dual core Pentium I3 laptop and on most occasions, the time taken to process the task (calculate 10 million square roots) using parallel processing is around 50% of that taken using sequential processing which, of course, is what you'd expect:

Imports System

Imports System.Diagnostics

Imports System.Threading.Tasks

Module Test

Sub Main()

' find square roots of first 10 million non-negative integers, store in an array

' and find the average to check that the results are the same

' for both sequential and parallel processing

Dim roots(10000000) As Double

Dim watch As StopWatch

' first do it using a sequential for statement

watch = StopWatch.StartNew()

For x As Integer = 0 To 9999999

roots(x) = Math.Sqrt(x)

watch.Stop()

Console.WriteLine("Time taken by sequential for = {0} ms", watch.ElapsedMilliseconds)

Dim Average As Double = AverageRoot(roots)

Console.WriteLine("Average using sequential for = {0}", Average)

' now do it using a parallel for statement

Console.WriteLine()

Array.Clear(roots, 0, roots.Length)

watch = StopWatch.StartNew()

Parallel.For(0, 10000000, Sub(x) roots(x) = Math.Sqrt(x))

watch.Stop()

Console.WriteLine("Time taken by parallel for = {0} ms", watch.ElapsedMilliseconds)

Average = AverageRoot(roots)

Console.WriteLine("Average using sequential for = {0}", Average)

Console.ReadKey

End Sub

Function AverageRoot(ByVal roots As Double()) As Double

Dim Total As Double = 0

For Each d As Double in roots

Total += d

Return Total / roots.Length

End Function

End Module

The overload of the Parallel.For method I'm using takes 3 arguments:

1. A starting value (inclusive).

2. An ending value (exclusive).

3. A lambda expression which calculates the root and stores it in the array.

As you probably know, some iterations will be allocated to a thread running on one core and the other iterations to a thread running on the other core. As only one thread can deal with a particular iteration, there's no chance of a collision when writing the result to the array and a lock is not therefore needed.

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Vulpes14y ago
Here's a simple example I've devised to illustrate Parallel.For and to compare it with the 'normal' sequential For statement.

On my dual core Pentium I3 laptop and on most occasions, the time taken to process the task (calculate 10 million square roots) using parallel processing is around 50% of that taken using sequential processing which, of course, is what you'd expect:

Imports System
Imports System.Diagnostics
Imports System.Threading.Tasks

Module Test

Sub Main()

' find square roots of first 10 million non-negative integers, store in an array
' and find the average to check that the results are the same
' for both sequential and parallel processing

Dim roots(10000000) As Double
Dim watch As StopWatch

' first do it using a sequential for statement

watch = StopWatch.StartNew()

For x As Integer = 0 To 9999999
roots(x) = Math.Sqrt(x)
Next

watch.Stop()

Console.WriteLine("Time taken by sequential for = {0} ms", watch.ElapsedMilliseconds)
Dim Average As Double = AverageRoot(roots)
Console.WriteLine("Average using sequential for = {0}", Average)

' now do it using a parallel for statement

Console.WriteLine()
Array.Clear(roots, 0, roots.Length)

watch = StopWatch.StartNew()

Parallel.For(0, 10000000, Sub(x) roots(x) = Math.Sqrt(x))

watch.Stop()

Console.WriteLine("Time taken by parallel for = {0} ms", watch.ElapsedMilliseconds)
Average = AverageRoot(roots)
Console.WriteLine("Average using sequential for = {0}", Average)

Console.ReadKey

End Sub

Function AverageRoot(ByVal roots As Double()) As Double

Dim Total As Double = 0
For Each d As Double in roots
Total += d
Next
Return Total / roots.Length

End Function

End Module

The overload of the Parallel.For method I'm using takes 3 arguments:

1. A starting value (inclusive).

2. An ending value (exclusive).

3. A lambda expression which calculates the root and stores it in the array.

As you probably know, some iterations will be allocated to a thread running on one core and the other iterations to a thread running on the other core. As only one thread can deal with a particular iteration, there's no chance of a collision when writing the result to the array and a lock is not therefore needed.