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10752

answers:

5

When writing some vb code I tripped and I'm still left wondering why. I was ORing the case expectation yet a value lieing within this range didn't warrant a match; why not?

Example Code:

    Select Case 2
        Case 0
            ''// Some logic
        Case 1
            ''// Some other logic
        Case 2 Or 3
            Console.WriteLine("hit")
     End Select

With the above I would naturally assume that "hit" would be printed, but that's not the case.

A: 

This will allow you to perform "something" in the case of 0, "something else" in the case of 1, "hit" in the case of 2 or 3 or "hit else" otherwise.

Select Case 2
    Case 0
        Console.WriteLine("something")
    Case 1
        Console.WriteLine("something else")
    Case Is 2 To 3
        Console.WriteLine("hit")
    Else
        Console.WriteLine("hit else")
 End Select
achinda99
+3  A: 

Edit: It appears I was wrong in assuming that VB.NET doesn't allow Case ORing. I was thinking in C# and IL and it appears I was wrong.

However, as someone pointed out, the reason your code did not work was because Case 2 Or 3 was evaluating 2 Or 3 as a bitwise or and hence evaluating to Case 3.

For clarification:


       2 binary = 0000 0010
       3 binary = 0000 0011
  2 Or 3 binary = 0000 0011 (= 3)

  Select Case 2
     Case 0            '--> no match

     Case 1            '--> no match

     Case 2 Or 3       '(equivalent to Case 3  --> no match)
   End Select

However, I feel that I should point out that for the sake of performance, one should not use such constructs. When the compiler encounters Select statements (switch in C#) it will try to compile them using lookup tables and the switch MSIL instruction but in the case where you have something like Case 1,2,11,55 the compiler will not be able to convert that to a lookup table and it will have to use a series of compares (which is like using If.. Else).

The point is that in order to really take advantage of the Select statement, the cases should be designed with that in mind. Otherwise, the only benefit is code readability.

A well designed switch is an O(1) operation whereas an poorly designed one (which is equivalent to a series of If..Then..Else statements) is an O(n) operation.

Thanks, that allows my mind to rest now. Reading the code logically didn't give me any hints that I was doing something wrong.
vanslly
I'm glad you were able to find the answer to the problem you were looking to solve and I hope the clarifications I added to my answer will help you craft better Select statements :)
+10  A: 

Use the comma operator to delimit case statements

Select Case 2
    Case 0,1,2,3
        Console.WriteLine("hit")
 End Select
JaredPar
my intent is not for case 0 and 1 to fall through. But the comma is what I was after.
vanslly
@downvoter, care to explain?
JaredPar
-1, because it does not answer the question as stated. It rather gives an alternative solution to the one attempted by the questioner. The question is: how does VB interpret the OR operator in a Select statement? Otherwise, the question must be edited to match the answer. But I would not suggest that, because @JohnT gave the correct answer.
Geoffrey Van Wyk
@Geoffrey, umm, the OP both accepted my answer and acknowledged it was what they were after.
JaredPar
@JaredPar, I know. @vanslly must restate the question, but then @JohnT, who answered correctly, loses the points.
Geoffrey Van Wyk
+9  A: 

JaredPar has it right but you can also use the To construct

Select Case 2
    Case 0,1
    Case 2 To 3
        Console.WriteLine("Hit")
End Select

This would be 0 or 1 do nothing, 2 or 3 print Hit...The To construct is a range...

Here's the MSDN

Jason Punyon
True, and I was using the To keyword, but it's slightly hacky because if say 2 and 3 were Enum vals and the enum gets refactored without taking this use into consideration - which if it's in some class deep in the bowels of a large system - then you have some unexpected breaks ;)
vanslly
+4  A: 

As Jared* said, you need to use the comma operator to delimit case statements.

The or you were doing is a bitwise OR, resulting in it being "3". Amusingly, "2 AND 3" would probably have worked for your specific case.

*Can't vote up with my puny rep, sorry!

John T