BitConverter.ToUInt64 Question

[george r smith]

Hi,

I created a byte array
--byte[] aByteArray = new byte[64];
and set aByteArray[0] and aByteArray[16] to '1'.

Because I want to create a ulong (UInt64) with these values I do:
UInt64 bb;
bb = BitConverter.To UInt64(aByteArr ay,0);
Console.WriteLi ne(bb);

I get a 1. So I am only getting the first eight bytes of aByteArray. This is
what
the docs say. So how can I get these values into a ulong and back again into
an array.

Why would they have BitConverter only deal with 8 bytes. I am trying hard to
understand
BitArray class, the BitConverter class and the interplay between them.

thanks
grs

[Nicholas Paldino [.NET/C# MVP]]

Re: BitConverter.To UInt64 Question

George,

This behavior is correct. Since a byte is 8 bits, and an Int64 is 64
bits, you need 8 bytes to get the representation of an Int64 (or ulong).

Now, if you are using a BitArray, then you are actually working with the
individual bits, not bytes, and you will need 64 entries to get an Int64
from that class.

Hope this helps.


--
- Nicholas Paldino [.NET/C# MVP]
- mvp@spam.guard. caspershouse.co m

"george r smith" <gsmith@budgete xt.com> wrote in message
news:O1IoptqsDH A.536@tk2msftng p13.phx.gbl...[color=blue]
> Hi,
>
> I created a byte array
> --byte[] aByteArray = new byte[64];
> and set aByteArray[0] and aByteArray[16] to '1'.
>
> Because I want to create a ulong (UInt64) with these values I do:
> UInt64 bb;
> bb = BitConverter.To UInt64(aByteArr ay,0);
> Console.WriteLi ne(bb);
>
> I get a 1. So I am only getting the first eight bytes of aByteArray. This[/color]
is[color=blue]
> what
> the docs say. So how can I get these values into a ulong and back again[/color]
into[color=blue]
> an array.
>
> Why would they have BitConverter only deal with 8 bytes. I am trying hard[/color]
to[color=blue]
> understand
> BitArray class, the BitConverter class and the interplay between them.
>
> thanks
> grs
>
>[/color]

[Jon Skeet [C# MVP]]

Re: BitConverter.To UInt64 Question

george r smith <gsmith@budgete xt.com> wrote:[color=blue]
> I created a byte array
> --byte[] aByteArray = new byte[64];
> and set aByteArray[0] and aByteArray[16] to '1'.
>
> Because I want to create a ulong (UInt64) with these values I do:
> UInt64 bb;
> bb = BitConverter.To UInt64(aByteArr ay,0);
> Console.WriteLi ne(bb);
>
> I get a 1. So I am only getting the first eight bytes of aByteArray. This is
> what the docs say. So how can I get these values into a ulong and back again into
> an array.[/color]

How do you expect to get 64 bytes into a ulong, when a ulong is only 8
bytes (64 *bits*) long?
[color=blue]
> Why would they have BitConverter only deal with 8 bytes.[/color]

Because longs and ulongs are only 8 bytes long.
[color=blue]
> I am trying hard to understand
> BitArray class, the BitConverter class and the interplay between them.[/color]

I'm not sure where BitArray comes in here.

Take a step back - what do you want to do, exactly?

--
Jon Skeet - <skeet@pobox.co m>

http://www.pobox.com/~skeet

If replying to the group, please do not mail me too

[george r smith]

Re: BitConverter.To UInt64 Question

Jon,

What I am trying to is two things.
1. Create a chess bitboard of 64 bits stored as a ulong.
2. Create an array (BitArray or regular array) of same size for ease of
debugging and display.
For this I want to capability of converting back and forth.

Thanks.
grs

[Jon Skeet [C# MVP]]

Re: BitConverter.To UInt64 Question

george r smith <gsmith@budgete xt.com> wrote:[color=blue]
> What I am trying to is two things.
> 1. Create a chess bitboard of 64 bits stored as a ulong.
> 2. Create an array (BitArray or regular array) of same size for ease of
> debugging and display.
> For this I want to capability of converting back and forth.[/color]

I'd suggest you just abstract it to a class which only keeps it in a
long, with properties which just manipulate the bits within the long.
For the sake of debugging you could just write a method which converted
it to a string representation. You don't usually need to change things
during debugging, just examine them (IME).

--
Jon Skeet - <skeet@pobox.co m>

http://www.pobox.com/~skeet

If replying to the group, please do not mail me too

[Jeffrey Tan[MSFT]]

Re: BitConverter.To UInt64 Question


Hi George,

Unsigned long need 8 bytes(64bits) to store its value.
I think you can first create a ulong variable. Then you do like below to
convert this ulong into a bitarray:
UInt64 bb=/*value*/;
byte [] storearr=BitCon verter.GetBytes (bb);
BitArray ba=new BitArray(storea rr);

Then you can use System.Collecti ons.IEnumerator to enumerate every bits of
the bitarray.
For more information, please refer to:

http://msdn.microsoft.com/library/default.asp?url=/library/en-us/cpref/html/

frlrfsystemcoll ectionsbitarray classtopic.asp

Hope this helps,
Best regards,
Jeffrey Tan
Microsoft Online Partner Support
Get Secure! - www.microsoft.com/security
This posting is provided "as is" with no warranties and confers no rights.