IsLittleEndian field reports false, but it must be Little-Endian?

问题

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...

回答1:

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.

回答2:

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.

回答3:

How are you checking it?

For example, run this short console app:

using System;

public class Test
{
    static void Main()
    {
        Console.WriteLine(BitConverter.IsLittleEndian);
    }
}

What does that print? Could you give details about what hardware and OS you're using?

回答4:

Like Marc says it seems to be a timing bug. Using the VS debugger to "peek" at the value at startup it returns false, but if I print it in a messagebox it returns true.

来源：https://stackoverflow.com/questions/2023672/islittleendian-field-reports-false-but-it-must-be-little-endian