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Im running on a Intel computer (Win7 64-bit) and according to what I read Intel is using Little-Endian. I try this out in C# with the following code:

byte[] b2 = new byte[] { 0, 1 };
short b2short = BitConverter.ToInt16(b2, 0);
and b2short == 256 as expected from a Little-Endian. 
Then I read that in .NET, the BitConverter.IsLittleEndian should reflect what endian the system is using, and when I check the variable in Visual Studio it reports false, ie it is NOT Little-Endian.
Does that have anything to do with 64-bit OS? Any ideas?
EDIT:
My collegue, who sits across from me, did the same test (Win Vista 32-bit) and got the same results
EDIT 2:
This is REALLY strange. Whenever I run the code, and break after the BitConverter did its thing, the IsLittleEndian == false.
BUT, if I add the line  Console.WriteLine(BitConverter.IsLittleEndian); afterwards it is TRUE:
byte[] b2 = new byte[] { 0, 1 };
short b2short = BitConverter.ToInt16(b2, 0);
Console.WriteLine(BitConverter.IsLittleEndian);
// Now the IsLittleEndian is true
But once I remove the Console.WriteLine it is false again.
I can also add that even if I break on the "Console.WriteLine" the IsLittleEndian == true, but if I remove the line altogether it is false.
EDIT 3:
As Mark Gravell pointed out, this must be some timing-bug. If I use the variable BitConverter.IsLittleEndian it is initialized, if I dont (and look on it when breaking) its not initialized and thus false...
                Which intel is it? Could it be IA64? Although I've been assured that even on IA64 it might use LE.
– Marc Gravell
                Jan 7, 2010 at 21:16
I wonder if this is a timing bug, perhaps related to "beforefieldinit"... how are you looking at the value? It is possible that the type-initializer (for BitConverter) isn't getting triggered by the VS debugger (which is peeking under the covers, so to speak). Especially since false is the default value for a field...
The IsLittleEndian static field is set in the static constructor; and the time that an initializer executes is... very hard to predict. If you are using a debugger, all bets are off. The only way to reliably check the value of this field is via code (when the CLR will run the initializer at some point before it is required):
bool isLittleEndian = BitConverter.IsLittleEndian;
Don't trust the debugger / watch windows etc.
                Im am looking at it by breaking the code or Console.Writeline. That might be it, it isnt initialized when breaking the code, but it is initialized when I am using it in code, like Console.WriteLine.
– Ted
                Jan 7, 2010 at 21:33
                Since he says it is false if he removes the line altogether, this leads me to think that he has put the breakpoint on the line before, which would trigger before BitConverter is used. I wonder what would happen if he just added another line after his call to ToInt16, which did nothing related to BitConverter, and breaked on that line instead, would it then show true?
– Lasse V. Karlsen
                Jan 7, 2010 at 21:34
                Lasse: If the BitConverter.IsLittleEndian is used (like the Console.WriteLine) it is initializeed even before the line is executed. This, it is true even before BitConverter is used at all.
– Ted
                Jan 7, 2010 at 21:39
                For extra fun: I've just discovered that this readonly property is in fact read/write through the visual studio "Immediate Window". The value you set gets reported back in the immediate window and by tooltips etc, but not if you access the flag in code.  The value you set also gets ignored when you actually do a conversion.
– Nigel Hawkins
                Sep 6, 2011 at 10:36
Raymond Chen gives an expanded answer to the question here (mirror, with better formatting here).
The gist of it is that:
  
Reading a member from the debugger does not execute the code to initialize that member.
So when you look at the field in Visual Studio, it will report false because the static initializer has not yet run.  However, if you use the field in code, the static initializer will run, causing the field to return the actual correct value.
                Jon: please read my "EDIT 2" above as an answer to your question. The behaviour is very strange...
– Ted
                Jan 7, 2010 at 21:30
        
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