Wednesday, March 03, 2010
Module RandomCrash

    Sub Main()
        Try
            Throw New Exception
        Catch ex As Exception When DateTime.Now.Ticks Mod 2 = 0

        End Try
    End Sub

End Module
posted on Wednesday, March 03, 2010 1:04:58 PM (Pacific Standard Time, UTC-08:00)  #    Comments [3]

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 Saturday, October 24, 2009

Now that Visual Studio 2010 Beta 2 is finally out the door, I’ve had a bit more time to spend coding on some of my personal projects. Yesterday, I happened upon a cool trick while using the new Generate from Usage feature. It was so helpful to me that I thought others might benefit, so I’m sharing it here.

The Anonymous Type Problem

When you need to project some data from a LINQ expression, anonymous types can be enormously convenient.

C# Query

Because anonymous types are… well… anonymous, they don’t have names that can be expressed in code. This is problematic if you want to expose an anonymous type as the return type of a function. I have run into this problem many times. When refactoring code, it’s easy to get into a situation like the one below.

Broken C# Function

So, how can you get around this problem? Well, there a few possibilities.

  1. You could replace ??? with object and use reflection to get at the properties. (Yuck!)
  2. You could make the function generic and add a parameter to “mumble” the anonymous type.1 (Awkward!)
  3. Assuming C# 4.0, you could replace ??? with dynamic.2 (No compiler errors!)

Because none of these solutions is particularly savory, most of us are forced to create a new named type to replace the anonymous type. Thankfully, there are some fantastic third-party refactoring tools out there that can automate this tedious process, but if you don’t use one of these tools you’re stuck writing the code by hand.

Actually, no, that’s not quite true.

Generate from Usage to The Rescue!

In Visual Studio 2010, the new Generate from Usage feature makes the task of coding up new classes a snap! Just type a new name for the anonymous type in the editor, making the code look like a type constructor followed by an object initializer. Then, press Ctrl+. to expand the smart tag that immediately appears and choose the first suggestion to generate a new class.

Generate Class

Next, expand each smart tag in the object initializer to generate each property.

Generate Property

When you’re finished, you should have a brand new class containing each property, declared as auto-implemented properties. Cool!

Generated Class

For Visual Basic coders, Generate from Usage is even easier. Let’s start with the same LINQ expression in VB. (Notice the lack of the “_” line continuation characters. Hooray for VB10 implicit line continuation!)

VB Query

Just like before, type the name of the new type that you wish to generate and press Ctrl+. to expand the smart tag. After choosing the first suggestion from the smart tag, you’re finished. The VB Generate Class feature will drill into the object initializer and generate all of the necessary properties at the same time that the class is generated.

VB Generate Class

Wrapping Up

Of course, this technique is not without flaws.

  • The resulting type is not immutable like the anonymous type that you’re replacing. To address this, you can easily modify the generated properties to be read-only.
  • The new type does not have the same structural equality semantics that anonymous types have. In practice, I’ve rarely run into an bug caused by anonymous type structural equality, but if this is a concern for you, use one of the excellent third-party tools that account for these differences.

 

1See Wes Dyer's excellent article for an example of this clever trick.
2Check out Bill Wagner's post for details.

posted on Saturday, October 24, 2009 10:01:40 AM (Pacific Standard Time, UTC-08:00)  #    Comments [5]

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 Thursday, June 25, 2009

The other day, I caught a quick snapshot of Bethany playing with her favorite new toy:

 

A special thanks to my friend Joseph Hill for providing her favorite monkey!

posted on Thursday, June 25, 2009 8:07:05 PM (Pacific Standard Time, UTC-08:00)  #    Comments [2]

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 Thursday, May 14, 2009

Last October, DevExpress released a massively powerful FREE tool specifically for C# developers called CodeRush Xpress. Today, in partnership with Microsoft, DevExpress has outdone themselves with a brand new version of CodeRush Xpress containing full support for both C# and Visual Basic. This release merges CodeRush Xpress with Refactor! for Visual Basic, creating one giant über-tool with an amazing features:

  • Duplicate Line
  • Highlight All References
  • Increase or Reduce Selection
  • Smart Clipboard Operations
  • Generate from Using (TDD)
  • Quick Navigation Window
  • Quick File Navigation
  • 60+ Refactorings!

Get it while it’s hot! You can download your FREE copy of CodeRush Xpress for Visual Studio 2008 from http://devexpress.com/crx.

posted on Thursday, May 14, 2009 4:59:29 PM (Pacific Standard Time, UTC-08:00)  #    Comments [0]

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 Tuesday, March 17, 2009

I’m back again with another post in my Yet Another Project Euler Series. As a reminder, my approach to these problems is to try to find the most beautiful solution that I can using F#. While performance is an important, I’m looking for the most elegant solutions that I can find.

Project Euler problem ten is another dealing prime numbers.

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

As you can guess, there’s not too much to this problem. In fact, we’ll declare just one simple supporting function.

let lessThan y x = x < y

Using the above function and our existing prime number generator, our solution is short and simple.

primes
  |> Seq.take_while (lessThan 2000000L)
  |> Seq.sum

I feel a little guilty that this post is so short, but that’s the benefit of spending two whole blog posts working out a prime number generator for problem seven.

posted on Tuesday, March 17, 2009 2:13:48 AM (Pacific Standard Time, UTC-08:00)  #    Comments [4]

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 Saturday, March 14, 2009

In Visual Basic .NET, there are several cases in which the statement completion list will present the user with a list of values rather than the standard completion set. Most often, this occurs when assigning to a variable of one of the common System.Drawing types, Color, Brush or Pen.

Value Completion List

At first glance, the screenshot above might seem as if the Visual Basic IDE has hard-coded a set of values into IntelliSense, but that’s not the case. In fact, this is caused by a seldom-used feature of XML Documentation that is supported by Visual Basic .NET, but isn’t currently supported by C#1. By cracking open the the XML Documentation file for System.Drawing.dll (located at C:\Windows\Microsoft.NET\Framework\v2.0.50727\en\System.Drawing.xml on my machine), we’ll see a curious XML tag on the System.Drawing.Color definition.

<member name="T:System.Drawing.Color">
  <
summary>Represents an ARGB (alpha, red, green, blue) color.</summary>
  <
filterpriority>1</filterpriority>
  <completionlist cref="T:System.Drawing.Color" />
</
member>

The highlighted completionlist tag above is used by Visual Basic to populate the completion list with the public shared2 fields and properties from the specified class or module. In this particular case, the XML documentation causes Visual Basic to populate the completion list with the public shared properties of System.Drawing.Color.

Don’t believe me? Just comment out the System.Drawing.Color completionlist tag above, save and restart Visual Studio to see how this influences the statement completion list.

Standard Completion List

Many of you are probably thinking, “so, is this just a nifty implementation detail, or is something I can actually use?” The answer is, yes, this something you can use today to customize Visual Basic’s statement completion. The code below demonstrates how the completionlist tag can be used.

''' <completionlist cref="CommonOperations"/>
Public Class Operation
     Private ReadOnly _execute As Func(Of Integer, Integer, Integer)

     Public Sub New(ByVal execute As Func(Of Integer, Integer, Integer))
         _execute = execute
     End Sub

     Public Function
Execute(ByVal arg1 As Integer, ByVal arg2 As Integer) As Integer
         Return
_execute(arg1, arg2)
     End Function
End Class

Public NotInheritable Class
CommonOperations
     Public Shared ReadOnly Add = New Operation(Function(x, y) x + y)
     Public Shared ReadOnly Subtract = New Operation(Function(x, y) x - y)
     Public Shared ReadOnly Multiply = New Operation(Function(x, y) x * y)
     Public Shared ReadOnly Divide = New Operation(Function(x, y) x / y)
End Class

Visual Basic will automatically pick up the completionlist tag in the code above and use it to populate the completion list like so.

Custom Value Completion List

While a bit limited, it’s pretty easy to customize the statement completion list experience for Visual Basic to make certain types of APIs more discoverable. It’s as simple as a single XML tag.

1Kevin Pilch-Bisson (C# IDE Developer Lead and swell guy) has a clever idea for supporting the completionlist tag in the C# statement completion list while staying true to the C# IntelliSense model.
2The VB "Shared" keyword = static in C#.

posted on Saturday, March 14, 2009 12:23:10 PM (Pacific Standard Time, UTC-08:00)  #    Comments [4]

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 Monday, March 09, 2009

Welcome back F# junkies! In this installment of YAPES1, we’re tackling Project Euler problem nine.

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a2 + b2 = c2

For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.

Like we’ve done before, we’ll break this problem down into steps:

  1. Generate a stream of Pythagorean triples.
  2. Find the Pythagorean triple whose sum is equal to 1000.
  3. Calculate the product of that triple.

Sound good?

The first order of business is to determine how to generate Pythagorean triples. It turns out that there are many ways to do this. However, to keep it simple, we will use Euclid’s classic formula:

Given a pair of positive integers m and n with m > n

a = 2mn, b = m2n2, c = m2 + n2

Truthfully, this formula doesn’t actually produce every possible Pythagorean triple. However, since the solution is within the set that it generates, that really isn’t a problem.

NeRd Note
To produce every Pythagorean triple, we would have to introduce an additional parameter, k, into the formula.
a = 2kmn, b = k(m2n2), c = k(m2 + n2)
This is left as an exercise for you, dear reader!

First, we’ll declare an F# function that calculates a Pythagorean triple using Euclid’s formula.

let getTriple m n =
  let a = 2*m*n
  let b = m*m - n*n
  let c = m*m + n*n
  a,b,c

Using the above function, it’s easy to produce a stream of Pythagorean triples. Our strategy will be to start with the pair, m = 2 and n = 1. To generate the next pair, we’ll first check if n + 1 is less than m. If it is, we’ll use m and n + 1 as the next pair, otherwise we’ll use m + 1 and 1. Using Seq.unfold, we can use this algorithm to produce an infinite stream of Pythagorean triples:

let triples =
  Seq.unfold (fun (m,n) ->
                let res = getTriple m n
                let next = if n+1 < m then m,n+1 else m+1,1
                Some(res, next))
             (2,1)

With our generator in place, we can easily solve the problem.

triples
  |> Seq.find (fun (a,b,c) –> a+b+c = 1000)
  |> (fun (a,b,c) –> a*b*c)

While the above solution works well, we can abstract further to produce a more elegant solution. First, we’ll declare three functions to handle the three operations that we’re performing explicitly, sum, product and equals.

let sum (a,b,c) = a+b+c
let product (a,b,c) = a*b*c
let equals x y = x = y

Now, we can restate our solution using these functions with a bit of function composition to produce a truly beautiful solution.

triples
  |> Seq.find (sum >> equals 1000)
  |> product

As we’ve seen once before, function composition is beautiful.

1Yet Another Project Euler Series

posted on Monday, March 09, 2009 7:05:47 AM (Pacific Standard Time, UTC-08:00)  #    Comments [1]

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 Saturday, March 07, 2009

Welcome coding ninjas!1 Continuing with Yet Another Project Euler Series (YAPES™), we come upon problem eight.

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Being one of the earlier Project Euler problems, this is pretty trivial to solve. We’ll approach the problem by first breaking it into a set of simple steps:

  1. Transform the 1000-digit number into a list of digits.
  2. Transform the list of digits into a list of quintuplets—that is, a list containing tuples of each five consecutive elements.
  3. Transform the list of quintuplets into a list containing the product of each.
  4. Find the maximum element in the list.

Some or all of the steps above could be combined to produce a more optimal solution. However, the beauty of F#—and functional programming in general—is the ability to easily break a problem like this into a series of data transformations that run quickly enough for most situations. When a problem is broken down into simple operations, it is easier to optimize in other ways (e.g. parallelization).

The first step is to transform the 1000-digit number into a list of digits. To facilitate working with such a large number in code, we’ll represent it as a multi-line string. Like before, we’ll break the problem of converting a string into a list of digits into a set of simple steps:

  1. Transform the string into a list of chars.
  2. Because this was a multi-line string, some of the chars will be line breaks and whitespace. So, we’ll filter the list to include only the chars that represent digits.
  3. Transform the list of digit chars into a list containing the numerical value of each.

Let’s define a few library functions to to help with each step above.

module String =
  /// Takes a string and produces a list of chars.
  let toChars (s : string) =
    s.ToCharArray() |> Array.to_list

module Char =
  /// Determines whether a char represents a digit.
  let isDigit (c : char) =
    System.Char.IsDigit(c)

  /// Converts a char representing a digit into its numerical value.
  let toNumber (c : char) =
    int c - int '0'

Armed with these, we can perform the 3 steps above in a declarative fashion.

let number =
  "73167176531330624919225119674426574742355349194934
   96983520312774506326239578318016984801869478851843
   85861560789112949495459501737958331952853208805511
   12540698747158523863050715693290963295227443043557
   66896648950445244523161731856403098711121722383113
   62229893423380308135336276614282806444486645238749
   30358907296290491560440772390713810515859307960866
   70172427121883998797908792274921901699720888093776
   65727333001053367881220235421809751254540594752243
   52584907711670556013604839586446706324415722155397
   53697817977846174064955149290862569321978468622482
   83972241375657056057490261407972968652414535100474
   82166370484403199890008895243450658541227588666881
   16427171479924442928230863465674813919123162824586
   17866458359124566529476545682848912883142607690042
   24219022671055626321111109370544217506941658960408
   07198403850962455444362981230987879927244284909188
   84580156166097919133875499200524063689912560717606
   05886116467109405077541002256983155200055935729725
   71636269561882670428252483600823257530420752963450"
   |> String.toChars
   |> List.filter Char.isDigit
   |> List.map Char.toNumber

The next step is to transform the list of digits into a list of quintuplets. To achieve this, we’ll write a simple recursive function.

let rec toQuintuplets l =
  match l with
  | x1::(x2::x3::x4::x5::_ as t) -> (x1,x2,x3,x4,x5)::(toQuintuplets t)
  | _ -> []

Notice that the first pattern match above is slightly more complicated as we bind the list starting with the next element in the list to t, to make the recursion cleaner.

NeRd Note
The toQuintuplets function defined above will work well for this particular problem but will stack overflow if passed too large of a list because it is head-recursive. There are a couple of ways to make it tail-recursive and avoid blowing the stack, but the most elegant is to employ a continuation.
let toQuintuplets l =
  let rec loop l cont =
    match l with
    | x1::(x2::x3::x4::x5::_ as t) -> loop t (fun l –> cont ((x1,x2,x3,x4,x5)::l))
    | _ -> cont []

  loop l (fun x -> x)
Of course, that unnecessarily complicates this solution by turning the logic inside out!

The last function we’ll define is a small helper to produce the product of all of the values in a quintuplet.

let product (x1,x2,x3,x4,x5) = x1*x2*x3*x4*x5

All of the pieces are now in place, and the steps can be combined to solve Project Euler problem eight.

number
  |> toQuintuplets
  |> List.map product
  |> List.max

It’s that simple!

1Or pirates.

posted on Saturday, March 07, 2009 1:09:11 PM (Pacific Standard Time, UTC-08:00)  #    Comments [2]

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