Re: Endian - C# Discussion Boards

31-May-07 5:42

[I thought I had found my answer earler so I had deleted the other post, but I didn't.]

I am working with data where records are stored in big and little endian.
For example a file has:
16 fields some are intergers and some a doubles and half are stored in big
and the other half in little.

I know that my system should be in a little endian state but..

How do I convert from big to little and little to big?

God Bless,
Jason

led mike31-May-07 6:02

led mike

31-May-07 6:02

The Windows Sockets library has built in functions or you could write your own using bitwise manipulations.

Luc Pattyn31-May-07 9:41

31-May-07 9:41

Hi,

I dont think mixing big-endian and little-endian data in a single file is a good idea.
If only one type of system is involved, convert the "wrong endian" data before storing
it; it different types of systems are involved, select one type, and let the systems
with the other convention do all the conversions every time.

Anyhow, you can change from one to the other with simple byte swapping
BTW the code is the same for either direction i.e. big2little and little2big
are no different.

The code can be based on byte extraction/insertion (using and, shift, or),
or on a union-like construct (using a struct with FieldOffsetAttribute).

Smile | :)

Luc Pattyn

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~~_{lost in transition}~~ 31-May-07 9:51

31-May-07 9:51

I agree. The problem is that the data files are created by another applications and I don't think I will be able to change their think on the subject.

Anyways I think I figure out a to flip them.

Correct me if I'm wrong if you have a 32bit integer (int32) it has 4 bytes.
So we could say a int32 bytes like this WXYZ flip to ZYXW.
If that is correct then my method works just fine.

static byte[] FlipBytes(byte[] bArray)
{
    byte[] bNewArray = new byte[bArray.Length];
    int j = 0;
    for (int i = bArray.Length - 1; i >= 0; i--)
    {
        bNewArray[j] = bArray[i];
        j++;
    }
    return bNewArray;
}

God Bless,
Jason

Luc Pattyn31-May-07 10:03

31-May-07 10:03

Yes, that is correct.
And a good way if you start off with a byte array.
If you start off with a long, an int, a short then it would be a pitty to first
create a byte array.

BTW you may not need the new byte array, you could swap bytes in situ:
it suffices to loop over half of the bytes and swap them (using one temporary byte).

int lenMinus1=bArray.Length-1;
for (int i = 0; i <= lenMinus1/2; i++) {
    int j=lenMinus1-i;
    byte temp=bArray[i];
    bArray[i]=bArray[j];
    bArray[j]=temp;

Luc Pattyn

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~~_{lost in transition}~~ 31-May-07 10:21

31-May-07 10:21

Thanks I got one right Smile | :)

Why would it be bad to create a byte array?

I think I might just do that.

Thanks for you input.

God Bless,
Jason

Luc Pattyn31-May-07 10:51

31-May-07 10:51

It is not bad, but why create and work with an object if there exists a simple
solution without such object.
Hence, if you start off with a byte array, operate on it; if you start off with
a long/int/short, operate on it. Example:

ushort val=0x1234;
val=EndianChanged(val);
...
public ushort EndianChanged(ushort val) {
    // right shifting an unsigned is OK !
    return val=(val>>8) | (val<<8); 
}

As you can guess, this method is much cheaper than the byte-array based solution,
it is less code and no new objects.

Of course, if you need this endian changing only a couple of times, it does not
matter how you do it. If you are going to read an entire binary program file,
maybe several MB in size, it may well matter.

Smile | :)

Luc Pattyn

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~~_{lost in transition}~~ 31-May-07 11:47

31-May-07 11:47

Okay I did a test to understand whats going on and verify the correct method to use
and this is what I came up with:

234                    /*original test number
Byte: 234              >
Byte: 0                >original test number
Byte: 0                >array of bytes
Byte: 0                */

Array Flip: -369098752 /*using the method I came up with
Byte: 0                > I got the signed 32bit integer
Byte: 0                > This is the 
Byte: 0                > array of bytes for that number
Byte: 234              */

Int Shift: 59904       /*using your shifting method
Byte: 0                > produced this unsigned 32bit integer
Byte: 234              > This is the 
Byte: 0                > array of bytes for that number
Byte: 0                */

IPAddress: -369098752  /*using the System.Net.IPAddress.HostToNetwork method I came up with
Byte: 0                > I got the signed 32bit integer
Byte: 0                > This is the 
Byte: 0                > array of bytes for that number
Byte: 234              */

The reason why I wanted to write my own method instead of using IPAddress is so that I
could understand whats going on and also I don't want to have to referance it.

I am not trying to be a pain I just want to know which is right and I'll use it.

God Bless,
Jason

Luc Pattyn31-May-07 12:00

31-May-07 12:00

Hi,

I am not sure I understand your last message completely, here are some thoughts:

- my val=(val>>8) | (val<<8) code is probably not completely accurate, I guess
it will need an overall cast to get rid of the intermediate data type (without
it, the compiler would complain; and when changing the return type, the value
would be wrong, since the left shift must loose its top byte).

- I dont see how it would ever turn (0,0,0,234) into (0,234,0,0).

- for testing purposes it is best to use a hex number that fills the variable,
so for an ushort I would use 0x1234, for a uint 0x12345678; this gives a maximum
of information while debugging

- yes you can use IPAddress.HostToNetworkOrder, but that is not what I would do
since the method's name does not reflect the intention. You could hide it
inside your own ChangeEndianness() method tho.

Smile | :)

Luc Pattyn

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~~_{lost in transition}~~ 1-Jun-07 4:58

1-Jun-07 4:58

Okay, its a new day lets look at this again.

using hex 0x12345678 = 305419896 that
gives us an array of bytes:
Byte: 120
Byte: 86
Byte: 52
Byte: 18

(If I am thinking correctly)Switching this from little endian to big endian
we would need to basically reverse the order of the bytes:
Byte: 18
Byte: 52
Byte: 86
Byte: 120
which is hex 0x78563412 = 2018915346

val = (val >> 8) | (val << 8) gives us:
Byte: 86
Byte: 124
Byte: 86
Byte: 52
which is hex 0x34567C56 = 878083158

Using your -val = (val >> 8) | (val << 8)- we are shifting bits, but I didn't think that is what
the problem is. For assumption was that big and little endian determined which way the bytes were
laid out. For example: 0x12345678 has 4 bytes(32-bits)-bytes:Z=120,Y=86,X=52,W=18. Now this is in
little endian basically because this machine uses little endian. Or in other words the bytes are read right to left in a least to greatest manner. Say that big endian is the opposite, then big read left to right in a least to greatest manner also. Big endian byte arrangement to 0x12345678 would be: W=18, X=52, Y=86, Z=120.

I am questioning things not to be a pain, but to make sure I understand. It seems to me that your are trying to shift bits when what is need is the bytes' order to be reversed. Yes/No Confused | :confused:

God Bless,
Jason

Luc Pattyn1-Jun-07 6:14

1-Jun-07 6:14

Hi,

the example I gave was handling unsigned shorts, these hold two bytes, hence the
formula included just one OR operator.
For (unsigned) ints, you need to swap 4 bytes, hence 3 OR operators; plus some
masking to avoid bit aliasing.

You could also do it pseudo-recursively like so:

// Swaps all bytes of a uint
public static uint Swap(uint val) {
return (uint)( swapLowerTwoBytes(val)<<16 | swapLowerTwoBytes(val>>16) );
}

// Swaps a ushort (but accepts and returns it as a uint)
public static uint swapLowerTwoBytes(uint val) {
ushort sval=(ushort)val; // throw away top bytes
return (ushort) ( (sval<8) | (sval>8) );
}

Both methods look very similar, I added the (uint) cast for clarity although
it is not needed; both (ushort) casts are necessary to throw away irrelevant bytes.

For signed int, it is similar, but be careful the sign bit gets replicated when
shifting right; either use an explicit AND, or cast to unsigned first.

Hope this helps.

BTW: when I told you to use a recognizable hex test value, the basic idea
is to do all printout in hex too !

Smile | :)

Luc Pattyn

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~~_{lost in transition}~~ 1-Jun-07 8:52

1-Jun-07 8:52

Thank you for the time you spent on this matter.

I think we've got it worked out. Smile | :)

God Bless,
Jason

Luc Pattyn1-Jun-07 10:08

1-Jun-07 10:08

You are welcome.

I hope you are sure you understand the problem and the solutions (with and
without byte array).

Smile | :)

Luc Pattyn

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~~_{lost in transition}~~ 31-May-07 11:52

31-May-07 11:52

Never mind I got it:

return val = (val >> 24) |(val << 24);

Yep your right that is better.

Thanks for the time and help.

God Bless,
Jason

Luc Pattyn31-May-07 12:05

31-May-07 12:05

?????????

what is val's type ? what did you do to my code, now it is completely wrong.

[1]

ushort val=0x1234 ?

val>>24=0

val<<24=0

so it returns 0

[2]

uint val=0x12345678 ?

val>>24=0x00000012

val<<24=0x78000000

so it returns 0x78000012 ????

?????????

Luc Pattyn

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~~_{lost in transition}~~ 31-May-07 9:43