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Converting to big -endianHello All - have a problem that I'm pulling my hair out on. I'm trying to
take a hex value and convert it to 16bit big-endian format. I can eaisly get it to little-endian format by convert.ToUInt16(&H600) - works like a champ. But there is no conver to UInt16 big-endian. Am I missing a conversion function somewhere or do I have to do some shifting and bitmasking to make this work? "Kenny" <Ke***@discussions.microsoft.com> wrote in message Any routine that converts endianness converts it both ways since it is just news:DCF4D673-2BFE-430A-BD1F-9A42F8E13B31@microsoft.com... > Hello All - have a problem that I'm pulling my hair out on. I'm trying to > take a hex value and convert it to 16bit big-endian format. I can eaisly > get > it to little-endian format by convert.ToUInt16(&H600) - works like a > champ. > But there is no conver to UInt16 big-endian. Am I missing a conversion > function somewhere or do I have to do some shifting and bitmasking to make > this work? a byte swap so you can use the same method for both. The "proper" way to do endianness is to have machine specific methods that either swap or don't depending on the architecture of the machine on which they are running - In this scenario you don't have to know your architecture or even think - you just call the method corresponding to the incoming/outgoing order on all values as they enter and leave your app. e.g. If my external interface is defined as BigEndian I write internalVal = BigEndian(externalVal); // incoming externalVal = BigEndian(internalVal); // outgoing and if little endian i write: internalVal = LittleEndian(externalVal); // incoming externalVal = LittleEndian(internalVal); // outgoing For any given architecture one of these methods will do nothing and the other will swap bytes but I don't have to know or care which it is which. As far as I know no major system does it this way but I came across it as macros in QNX/Momentics when I was writing embedded C++ and it certainly makes life much simpler. Of course as macros there is no cost in the no-op method which is not the case with C#. You could have the same effect by using something like: [Conditional("BigEndian")] void LittleEndian(ref int i) { i = Swap(i); } [Conditional("LittleEndian")] void BigEndian(ref int i) { i = Swap(i); } Of course I appreciate that all the readers of this thread are so embedded in the MS/intel world that this is a non-issue but personally I keep forgetting and I like the elegance of this approach. Create a byte array of the same size as the original value. Read the bytes
in the original value into the array in reverse order. Use the BitConverter class to convert the byte array to the return type. Piece of cake. -- Show quoteHTH, Kevin Spencer Microsoft MVP ..Net Developer Presuming that God is "only an idea" - Ideas exist. Therefore, God exists. "Kenny" <Ke***@discussions.microsoft.com> wrote in message news:DCF4D673-2BFE-430A-BD1F-9A42F8E13B31@microsoft.com... > Hello All - have a problem that I'm pulling my hair out on. I'm trying to > take a hex value and convert it to 16bit big-endian format. I can eaisly > get > it to little-endian format by convert.ToUInt16(&H600) - works like a > champ. > But there is no conver to UInt16 big-endian. Am I missing a conversion > function somewhere or do I have to do some shifting and bitmasking to make > this work? Tried that and it doesnt work... Problem is that bitconvert .ToUInt16
converts back to little-endian. I need to basically mimick the PERL pack function... If I do this in PERL - $header = pack("n", 0xD0CF); I get d0 cf if I $header = pack("v", 0xD0CF); I get cf d0 - in .net if I convert.tou16int(&HD0CF) I get as expected cf d0 - but every different conversion to big-endian I try I get reults like 35 0 34 0 36 .... Show quote "Kevin Spencer" wrote: > Create a byte array of the same size as the original value. Read the bytes > in the original value into the array in reverse order. Use the BitConverter > class to convert the byte array to the return type. Piece of cake. > > -- > HTH, > > Kevin Spencer > Microsoft MVP > ..Net Developer > > Presuming that God is "only an idea" - > Ideas exist. > Therefore, God exists. > > "Kenny" <Ke***@discussions.microsoft.com> wrote in message > news:DCF4D673-2BFE-430A-BD1F-9A42F8E13B31@microsoft.com... > > Hello All - have a problem that I'm pulling my hair out on. I'm trying to > > take a hex value and convert it to 16bit big-endian format. I can eaisly > > get > > it to little-endian format by convert.ToUInt16(&H600) - works like a > > champ. > > But there is no conver to UInt16 big-endian. Am I missing a conversion > > function somewhere or do I have to do some shifting and bitmasking to make > > this work? > > > There seems to be some misunderstanding going on.
1. BitConverter.ToUInt16 takes a byte array and an index not an int (hex or otherwise) 2. Convert.ToUint16 only does extension and signed/unsigned conversion. 3. The fact that the input value to any method is hexadecimal constant is irrelevant. IPAddress.HostToNetworkOrder(Int16 val) will do the byte swapping for you on a little endian machine (i.e. intel) and you an either use (unchecked) casts or the full ugliness of Convert.ToUInt16( IPAddress.HostToNetworkOrder( Convert.ToInt16( myUInt ) ) ); It seems to me that the BitConverter methods should be instance methods not static and there should be an instance for big endian and an instance for little endian. Show quote "Kenny" <Ke***@discussions.microsoft.com> wrote in message news:48FC4A2D-84C8-4459-87A7-05F1A6078429@microsoft.com... > Tried that and it doesnt work... Problem is that bitconvert .ToUInt16 > converts back to little-endian. I need to basically mimick the PERL pack > function... If I do this in PERL - $header = pack("n", 0xD0CF); I get d0 > cf > if I $header = pack("v", 0xD0CF); I get cf d0 - in .net if I > convert.tou16int(&HD0CF) I get as expected cf d0 - but every different > conversion to big-endian I try I get reults like 35 0 34 0 36 .... > > "Kevin Spencer" wrote: > >> Create a byte array of the same size as the original value. Read the >> bytes >> in the original value into the array in reverse order. Use the >> BitConverter >> class to convert the byte array to the return type. Piece of cake. >> >> -- >> HTH, >> >> Kevin Spencer >> Microsoft MVP >> ..Net Developer >> >> Presuming that God is "only an idea" - >> Ideas exist. >> Therefore, God exists. >> >> "Kenny" <Ke***@discussions.microsoft.com> wrote in message >> news:DCF4D673-2BFE-430A-BD1F-9A42F8E13B31@microsoft.com... >> > Hello All - have a problem that I'm pulling my hair out on. I'm trying >> > to >> > take a hex value and convert it to 16bit big-endian format. I can >> > eaisly >> > get >> > it to little-endian format by convert.ToUInt16(&H600) - works like a >> > champ. >> > But there is no conver to UInt16 big-endian. Am I missing a conversion >> > function somewhere or do I have to do some shifting and bitmasking to >> > make >> > this work? >> >> >> Nick.. yes you are correct... and thanks for the input.. I kept getting
convert.touint16 confused with bitconverter.touint16.. sublte distinction that can get overlooked.. I actually took Kevin's advice and put into an array an looped backwards then use the bitocnverter with the proper index. This whole process is so annoying all for a customer who wants a data file with one particular column in big-endian order.. why I have no idea... any way here is what I did to make it work. .any suggestions for makign better? Dim foo As Byte() = BitConverter.GetBytes(&H1AE1) Dim bar() As Byte = New Byte((foo.Length - 1)) {} Dim i As Integer = 0 Dim u As Integer = 0 For i = foo.Length - 1 To 0 Step -1 bar(u) = foo(i) u += 1 Next BitConverter.ToUInt16(bar, 2) this will mimick perl... pack("n",0x1AE1) Kevin Spencer.. r u the same Kevin Spencer from takempis.com? If so remember me I was one of your MVP a long time ago.. how have you been? Show quote "Nick Hounsome" wrote: > There seems to be some misunderstanding going on. > > 1. BitConverter.ToUInt16 takes a byte array and an index not an int (hex or > otherwise) > 2. Convert.ToUint16 only does extension and signed/unsigned conversion. > 3. The fact that the input value to any method is hexadecimal constant is > irrelevant. > > IPAddress.HostToNetworkOrder(Int16 val) will do the byte swapping for you on > a little endian machine (i.e. intel) > and you an either use (unchecked) casts or the full ugliness of > > Convert.ToUInt16( IPAddress.HostToNetworkOrder( Convert.ToInt16( > myUInt ) ) ); > > It seems to me that the BitConverter methods should be instance methods not > static and there should be an instance for big endian and an instance for > little endian. > > "Kenny" <Ke***@discussions.microsoft.com> wrote in message > news:48FC4A2D-84C8-4459-87A7-05F1A6078429@microsoft.com... > > Tried that and it doesnt work... Problem is that bitconvert .ToUInt16 > > converts back to little-endian. I need to basically mimick the PERL pack > > function... If I do this in PERL - $header = pack("n", 0xD0CF); I get d0 > > cf > > if I $header = pack("v", 0xD0CF); I get cf d0 - in .net if I > > convert.tou16int(&HD0CF) I get as expected cf d0 - but every different > > conversion to big-endian I try I get reults like 35 0 34 0 36 .... > > > > "Kevin Spencer" wrote: > > > >> Create a byte array of the same size as the original value. Read the > >> bytes > >> in the original value into the array in reverse order. Use the > >> BitConverter > >> class to convert the byte array to the return type. Piece of cake. > >> > >> -- > >> HTH, > >> > >> Kevin Spencer > >> Microsoft MVP > >> ..Net Developer > >> > >> Presuming that God is "only an idea" - > >> Ideas exist. > >> Therefore, God exists. > >> > >> "Kenny" <Ke***@discussions.microsoft.com> wrote in message > >> news:DCF4D673-2BFE-430A-BD1F-9A42F8E13B31@microsoft.com... > >> > Hello All - have a problem that I'm pulling my hair out on. I'm trying > >> > to > >> > take a hex value and convert it to 16bit big-endian format. I can > >> > eaisly > >> > get > >> > it to little-endian format by convert.ToUInt16(&H600) - works like a > >> > champ. > >> > But there is no conver to UInt16 big-endian. Am I missing a conversion > >> > function somewhere or do I have to do some shifting and bitmasking to > >> > make > >> > this work? > >> > >> > >> > > > Hi Kenny,
Your code looks pretty good. Yes, I'm the same guy, but it's been a long time, so I can't say specifically that I remember you. Still, it's good to hear from you again! -- Show quoteKevin Spencer Microsoft MVP ..Net Developer Presuming that God is "only an idea" - Ideas exist. Therefore, God exists. "Kenny" <Ke***@discussions.microsoft.com> wrote in message news:24F2911F-3587-466F-A35F-68B5585B259B@microsoft.com... > Nick.. yes you are correct... and thanks for the input.. I kept getting > convert.touint16 confused with bitconverter.touint16.. sublte distinction > that can get overlooked.. I actually took Kevin's advice and put into an > array an looped backwards then use the bitocnverter with the proper index. > This whole process is so annoying all for a customer who wants a data file > with one particular column in big-endian order.. why I have no idea... > any > way here is what I did to make it work. .any suggestions for makign > better? > > Dim foo As Byte() = BitConverter.GetBytes(&H1AE1) > Dim bar() As Byte = New Byte((foo.Length - 1)) {} > > Dim i As Integer = 0 > Dim u As Integer = 0 > For i = foo.Length - 1 To 0 Step -1 > bar(u) = foo(i) > u += 1 > Next > > BitConverter.ToUInt16(bar, 2) > > this will mimick perl... pack("n",0x1AE1) > > Kevin Spencer.. r u the same Kevin Spencer from takempis.com? If so > remember > me I was one of your MVP a long time ago.. how have you been? > > "Nick Hounsome" wrote: > >> There seems to be some misunderstanding going on. >> >> 1. BitConverter.ToUInt16 takes a byte array and an index not an int (hex >> or >> otherwise) >> 2. Convert.ToUint16 only does extension and signed/unsigned conversion. >> 3. The fact that the input value to any method is hexadecimal constant >> is >> irrelevant. >> >> IPAddress.HostToNetworkOrder(Int16 val) will do the byte swapping for you >> on >> a little endian machine (i.e. intel) >> and you an either use (unchecked) casts or the full ugliness of >> >> Convert.ToUInt16( IPAddress.HostToNetworkOrder( Convert.ToInt16( >> myUInt ) ) ); >> >> It seems to me that the BitConverter methods should be instance methods >> not >> static and there should be an instance for big endian and an instance for >> little endian. >> >> "Kenny" <Ke***@discussions.microsoft.com> wrote in message >> news:48FC4A2D-84C8-4459-87A7-05F1A6078429@microsoft.com... >> > Tried that and it doesnt work... Problem is that bitconvert .ToUInt16 >> > converts back to little-endian. I need to basically mimick the PERL >> > pack >> > function... If I do this in PERL - $header = pack("n", 0xD0CF); I get >> > d0 >> > cf >> > if I $header = pack("v", 0xD0CF); I get cf d0 - in .net if I >> > convert.tou16int(&HD0CF) I get as expected cf d0 - but every different >> > conversion to big-endian I try I get reults like 35 0 34 0 36 .... >> > >> > "Kevin Spencer" wrote: >> > >> >> Create a byte array of the same size as the original value. Read the >> >> bytes >> >> in the original value into the array in reverse order. Use the >> >> BitConverter >> >> class to convert the byte array to the return type. Piece of cake. >> >> >> >> -- >> >> HTH, >> >> >> >> Kevin Spencer >> >> Microsoft MVP >> >> ..Net Developer >> >> >> >> Presuming that God is "only an idea" - >> >> Ideas exist. >> >> Therefore, God exists. >> >> >> >> "Kenny" <Ke***@discussions.microsoft.com> wrote in message >> >> news:DCF4D673-2BFE-430A-BD1F-9A42F8E13B31@microsoft.com... >> >> > Hello All - have a problem that I'm pulling my hair out on. I'm >> >> > trying >> >> > to >> >> > take a hex value and convert it to 16bit big-endian format. I can >> >> > eaisly >> >> > get >> >> > it to little-endian format by convert.ToUInt16(&H600) - works like a >> >> > champ. >> >> > But there is no conver to UInt16 big-endian. Am I missing a >> >> > conversion >> >> > function somewhere or do I have to do some shifting and bitmasking >> >> > to >> >> > make >> >> > this work? >> >> >> >> >> >> >> >> >> But this is much slower than bit shuffling with HostToNetworkOrder or just
doing it yourself. Why would you bother splitting it into a byte array first? Even if it is already in a byte array it is simpler code to just convert it and then use HostToNetworkOrder to shufle the bits. Show quote "Kenny" <Ke***@discussions.microsoft.com> wrote in message news:24F2911F-3587-466F-A35F-68B5585B259B@microsoft.com... > Nick.. yes you are correct... and thanks for the input.. I kept getting > convert.touint16 confused with bitconverter.touint16.. sublte distinction > that can get overlooked.. I actually took Kevin's advice and put into an > array an looped backwards then use the bitocnverter with the proper index. > This whole process is so annoying all for a customer who wants a data file > with one particular column in big-endian order.. why I have no idea... > any > way here is what I did to make it work. .any suggestions for makign > better? > > Dim foo As Byte() = BitConverter.GetBytes(&H1AE1) > Dim bar() As Byte = New Byte((foo.Length - 1)) {} > > Dim i As Integer = 0 > Dim u As Integer = 0 > For i = foo.Length - 1 To 0 Step -1 > bar(u) = foo(i) > u += 1 > Next > > BitConverter.ToUInt16(bar, 2) > > this will mimick perl... pack("n",0x1AE1) > > Kevin Spencer.. r u the same Kevin Spencer from takempis.com? If so > remember > me I was one of your MVP a long time ago.. how have you been? > > "Nick Hounsome" wrote: > >> There seems to be some misunderstanding going on. >> >> 1. BitConverter.ToUInt16 takes a byte array and an index not an int (hex >> or >> otherwise) >> 2. Convert.ToUint16 only does extension and signed/unsigned conversion. >> 3. The fact that the input value to any method is hexadecimal constant >> is >> irrelevant. >> >> IPAddress.HostToNetworkOrder(Int16 val) will do the byte swapping for you >> on >> a little endian machine (i.e. intel) >> and you an either use (unchecked) casts or the full ugliness of >> >> Convert.ToUInt16( IPAddress.HostToNetworkOrder( Convert.ToInt16( >> myUInt ) ) ); >> >> It seems to me that the BitConverter methods should be instance methods >> not >> static and there should be an instance for big endian and an instance for >> little endian. >> >> "Kenny" <Ke***@discussions.microsoft.com> wrote in message >> news:48FC4A2D-84C8-4459-87A7-05F1A6078429@microsoft.com... >> > Tried that and it doesnt work... Problem is that bitconvert .ToUInt16 >> > converts back to little-endian. I need to basically mimick the PERL >> > pack >> > function... If I do this in PERL - $header = pack("n", 0xD0CF); I get >> > d0 >> > cf >> > if I $header = pack("v", 0xD0CF); I get cf d0 - in .net if I >> > convert.tou16int(&HD0CF) I get as expected cf d0 - but every different >> > conversion to big-endian I try I get reults like 35 0 34 0 36 .... >> > >> > "Kevin Spencer" wrote: >> > >> >> Create a byte array of the same size as the original value. Read the >> >> bytes >> >> in the original value into the array in reverse order. Use the >> >> BitConverter >> >> class to convert the byte array to the return type. Piece of cake. >> >> >> >> -- >> >> HTH, >> >> >> >> Kevin Spencer >> >> Microsoft MVP >> >> ..Net Developer >> >> >> >> Presuming that God is "only an idea" - >> >> Ideas exist. >> >> Therefore, God exists. >> >> >> >> "Kenny" <Ke***@discussions.microsoft.com> wrote in message >> >> news:DCF4D673-2BFE-430A-BD1F-9A42F8E13B31@microsoft.com... >> >> > Hello All - have a problem that I'm pulling my hair out on. I'm >> >> > trying >> >> > to >> >> > take a hex value and convert it to 16bit big-endian format. I can >> >> > eaisly >> >> > get >> >> > it to little-endian format by convert.ToUInt16(&H600) - works like a >> >> > champ. >> >> > But there is no conver to UInt16 big-endian. Am I missing a >> >> > conversion >> >> > function somewhere or do I have to do some shifting and bitmasking >> >> > to >> >> > make >> >> > this work? >> >> >> >> >> >> >> >> >> Hi Nick,
> But this is much slower than bit shuffling with HostToNetworkOrder or just I suppose it all depends on what you're doing. If you're working with a > doing it yourself. UInt16 in the first place, maybe. I'm not entirely sure that using HostToNetworkOrder is more efficient otherwise. It may simply be doing what you're talking about. After all, the bytes *do* have to be reversed. So, in your case, you're adding the step of converting the bytes to a UInt16 and then calling a function which converts the UInt16 to an array of bytes, reverses the order, and then converts the array of bytes back to a number. I suppose it is possible that it does it all in an unmanaged bitwise way under the hood, but I doubt it. In addition, this is fine and dandy if the data you're working with is confined to 16, 32, or 64-bit numbers. Otherwise, you will have to roll your own method, as I've done in my library, which works with a large variety of data types, for the purpose of reading TIFF files. -- Show quoteHTH, Kevin Spencer Microsoft MVP ..Net Developer Presuming that God is "only an idea" - Ideas exist. Therefore, God exists. "Nick Hounsome" <nh***@nickhounsome.me.uk> wrote in message news:_FPSf.208805$YJ4.89282@fe2.news.blueyonder.co.uk... > But this is much slower than bit shuffling with HostToNetworkOrder or just > doing it yourself. > > Why would you bother splitting it into a byte array first? > > Even if it is already in a byte array it is simpler code to just convert > it and then use HostToNetworkOrder to shufle the bits. > > "Kenny" <Ke***@discussions.microsoft.com> wrote in message > news:24F2911F-3587-466F-A35F-68B5585B259B@microsoft.com... >> Nick.. yes you are correct... and thanks for the input.. I kept getting >> convert.touint16 confused with bitconverter.touint16.. sublte distinction >> that can get overlooked.. I actually took Kevin's advice and put into an >> array an looped backwards then use the bitocnverter with the proper >> index. >> This whole process is so annoying all for a customer who wants a data >> file >> with one particular column in big-endian order.. why I have no idea... >> any >> way here is what I did to make it work. .any suggestions for makign >> better? >> >> Dim foo As Byte() = BitConverter.GetBytes(&H1AE1) >> Dim bar() As Byte = New Byte((foo.Length - 1)) {} >> >> Dim i As Integer = 0 >> Dim u As Integer = 0 >> For i = foo.Length - 1 To 0 Step -1 >> bar(u) = foo(i) >> u += 1 >> Next >> >> BitConverter.ToUInt16(bar, 2) >> >> this will mimick perl... pack("n",0x1AE1) >> >> Kevin Spencer.. r u the same Kevin Spencer from takempis.com? If so >> remember >> me I was one of your MVP a long time ago.. how have you been? >> >> "Nick Hounsome" wrote: >> >>> There seems to be some misunderstanding going on. >>> >>> 1. BitConverter.ToUInt16 takes a byte array and an index not an int (hex >>> or >>> otherwise) >>> 2. Convert.ToUint16 only does extension and signed/unsigned conversion. >>> 3. The fact that the input value to any method is hexadecimal constant >>> is >>> irrelevant. >>> >>> IPAddress.HostToNetworkOrder(Int16 val) will do the byte swapping for >>> you on >>> a little endian machine (i.e. intel) >>> and you an either use (unchecked) casts or the full ugliness of >>> >>> Convert.ToUInt16( IPAddress.HostToNetworkOrder( Convert.ToInt16( >>> myUInt ) ) ); >>> >>> It seems to me that the BitConverter methods should be instance methods >>> not >>> static and there should be an instance for big endian and an instance >>> for >>> little endian. >>> >>> "Kenny" <Ke***@discussions.microsoft.com> wrote in message >>> news:48FC4A2D-84C8-4459-87A7-05F1A6078429@microsoft.com... >>> > Tried that and it doesnt work... Problem is that bitconvert .ToUInt16 >>> > converts back to little-endian. I need to basically mimick the PERL >>> > pack >>> > function... If I do this in PERL - $header = pack("n", 0xD0CF); I get >>> > d0 >>> > cf >>> > if I $header = pack("v", 0xD0CF); I get cf d0 - in .net if I >>> > convert.tou16int(&HD0CF) I get as expected cf d0 - but every >>> > different >>> > conversion to big-endian I try I get reults like 35 0 34 0 36 .... >>> > >>> > "Kevin Spencer" wrote: >>> > >>> >> Create a byte array of the same size as the original value. Read the >>> >> bytes >>> >> in the original value into the array in reverse order. Use the >>> >> BitConverter >>> >> class to convert the byte array to the return type. Piece of cake. >>> >> >>> >> -- >>> >> HTH, >>> >> >>> >> Kevin Spencer >>> >> Microsoft MVP >>> >> ..Net Developer >>> >> >>> >> Presuming that God is "only an idea" - >>> >> Ideas exist. >>> >> Therefore, God exists. >>> >> >>> >> "Kenny" <Ke***@discussions.microsoft.com> wrote in message >>> >> news:DCF4D673-2BFE-430A-BD1F-9A42F8E13B31@microsoft.com... >>> >> > Hello All - have a problem that I'm pulling my hair out on. I'm >>> >> > trying >>> >> > to >>> >> > take a hex value and convert it to 16bit big-endian format. I can >>> >> > eaisly >>> >> > get >>> >> > it to little-endian format by convert.ToUInt16(&H600) - works like >>> >> > a >>> >> > champ. >>> >> > But there is no conver to UInt16 big-endian. Am I missing a >>> >> > conversion >>> >> > function somewhere or do I have to do some shifting and bitmasking >>> >> > to >>> >> > make >>> >> > this work? >>> >> >>> >> >>> >> >>> >>> >>> > > You lost me, Kenny. I have been using the function I described for quite
awhile without issues. The only difference is that I am reading from a stream. But I'm reversing the bytes and using a BitConverter to convert them back. Here's the acutal code I wrote: /// <summary acc="protected" type="System.UInt16"> /// Overloaded.Reads a <c>System.UInt16</c> value using Big-Endian /// or Little-Endian byte order. /// </summary> /// <param name="reader"><c>System.IO.BinaryReader</c> to read from.</param> /// <returns><c>System.UInt16</c> value.</returns> /// <remarks>Reads from the current position of the <paramref name="reader"/>.</remarks> protected ushort ReadUInt16(BinaryReader reader) { if (_ByteOrder == ByteOrder.LittleEndian) { return reader.ReadUInt16(); } else // Big-Endian { byte[] bytes = new byte[2]; bytes[1] = reader.ReadByte(); bytes[0] = reader.ReadByte(); return BitConverter.ToUInt16(bytes, 0); } } /// <summary acc="protected" type="System.UInt16"> /// Overloaded.Reads a <c>System.UInt16</c> value using Big-Endian /// or Little-Endian byte order. /// </summary> /// <param name="reader"><c>System.IO.BinaryReader</c> to read from.</param> /// <param name="Offset">Offset from beginning of file.</param> /// <returns><c>System.UInt16</c> value.</returns> protected ushort ReadUInt16(BinaryReader reader, long Offset) { reader.BaseStream.Seek(Offset, SeekOrigin.Begin); return ReadUInt16(reader); } -- Show quoteHTH, Kevin Spencer Microsoft MVP ..Net Developer Presuming that God is "only an idea" - Ideas exist. Therefore, God exists. "Kenny" <Ke***@discussions.microsoft.com> wrote in message news:48FC4A2D-84C8-4459-87A7-05F1A6078429@microsoft.com... > Tried that and it doesnt work... Problem is that bitconvert .ToUInt16 > converts back to little-endian. I need to basically mimick the PERL pack > function... If I do this in PERL - $header = pack("n", 0xD0CF); I get d0 > cf > if I $header = pack("v", 0xD0CF); I get cf d0 - in .net if I > convert.tou16int(&HD0CF) I get as expected cf d0 - but every different > conversion to big-endian I try I get reults like 35 0 34 0 36 .... > > "Kevin Spencer" wrote: > >> Create a byte array of the same size as the original value. Read the >> bytes >> in the original value into the array in reverse order. Use the >> BitConverter >> class to convert the byte array to the return type. Piece of cake. >> >> -- >> HTH, >> >> Kevin Spencer >> Microsoft MVP >> ..Net Developer >> >> Presuming that God is "only an idea" - >> Ideas exist. >> Therefore, God exists. >> >> "Kenny" <Ke***@discussions.microsoft.com> wrote in message >> news:DCF4D673-2BFE-430A-BD1F-9A42F8E13B31@microsoft.com... >> > Hello All - have a problem that I'm pulling my hair out on. I'm trying >> > to >> > take a hex value and convert it to 16bit big-endian format. I can >> > eaisly >> > get >> > it to little-endian format by convert.ToUInt16(&H600) - works like a >> > champ. >> > But there is no conver to UInt16 big-endian. Am I missing a conversion >> > function somewhere or do I have to do some shifting and bitmasking to >> > make >> > this work? >> >> >>  |
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