How to find machine is Big/little endian?

Hi,
Can any one tell me that how to find machine on which we are working i.e. Big/Little Endian in using java?
One of my friend told me that in C Programming we can find this by using following code:
#include <stdio.h>
#include <arpa/inet.h>
int main() {
    if (htons(1) == 1) puts("big endian");
    else puts("little endian");
    return 0;
}Can any one help me in solving this mystery?
Cheers,
Ram

java.nio.ByteOrder.nativeOrder();~D

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Check thie following link:
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* @author
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Little endian & big endian format system

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hi
chk this
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Effect
Data is read or written in a form which is compatible to BINARY MODE in Releases <= 4.6. This addition is primarily used to convert a file into the code page format specified already when it is opened. At runtime, the system uses the format of the system code page of the application server. The system saves the file then again in the code page specified. This procedure is important if data is exchanged between systems using different code pages. For more information, see READ DATASET and TRANSFER.
Notes
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These additions specify the byte sequence in which to store numbers (ABAP types I, F, and INT2) in the file.
These additions may only be used in combination with the additions IN LEGACY BINARY MODE and IN LEGACY TEXT MODE. If these are not specified, the system assumes that the byte sequence determined by the hardware of the application server is used in the file.
If the byte sequence specified differs from that determined by the hardware of the application server, READDATASET and TRANSFER make the corresponding conversions.
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on CODE PAGE cp
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This addition replaces the language element TRANSLATE ... CODE PAGE ... which must not be used in Unicode programs.
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Data is read or written in a form which is compatible to BINARY MODE in Releases <= 4.6. This addition is primarily used to convert a file into the code page format specified already when it is opened. At runtime, the system uses the format of the system code page of the application server. The system saves the file then again in the code page specified. This procedure is important if data is exchanged between systems using different code pages. For more information, see READ DATASET and TRANSFER.

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How do I swap 64-bit and 32-bit floats from little-endian to big-ending

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I know that if I were to read the little-endian float into the big-endian float type (float or double) that the format is pretty much lost (from little I understand about how floating point values are stored in memory).
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A 64-bit double is represented by the sign bit, an 11-bit (biased) exponent
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when stored to memory, as you observed.
A 32-bit float is represented by the sign bit, an 8-bit (biased) exponent
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As you can see, a 64-bit double is not merely a pair of 32-bit float.
You need to know exactly how the floating-point data was written:
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How to judge the platform is little or big endian ?

I have a c code, but anybody has a Java code?
Thanks.
#include <stdio.h>
//little-endian or big-endian
bool IsLittleEndian()
int i = 1;
char* p = (char*)&i;
return *p;
int main()
if(IsLittleEndian())
printf("LittleEndian\n");
else
printf("BigEndian\n");
return 0;
}

http://www.codeproject.com/cpp/endianness.asp
I got one, but how to translate into Java code.
Auto detecting the correct Endian format of a data file
Suppose you are developing a Windows application that imports Nuclear Magnetic Resonance (NMR) spectra. High resolution NMR files are generally recorded in Silicon or Sun Workstations but recently Windows or Linux based spectrometers are emerging as practical substitutes. It turns out that you will need to know in advance the Endian format of the file to parse correctly all the information. Here are some practical guidelines you can follow to decipher the correct Endianness of a data file:
Typically, the binary file includes a header with the information about the Endian format.
If the header is not present , you can guess the Endian format if you know the native format of the computer from which the file comes from. For instance, if the file was created in a Sun Workstation, the Endian format will most likely be Big-Endian.
If none of the above points apply, the Endian format can be determined by trial and error. For example, if after reading the file assuming one format, the spectrum does not make sense, you will know that you have to use the other format.
If the data points in the file are in floating point format (double), then the _isnan() function can be of some help to determine the Endian format. For example:
double dValue;
FILE* fp;
fread(&nValue, 1, sizeof(double), fp);
bool bByteSwap = _isnan(dValue) ? true : falseNote that this method does only guarantee that the byte swap operation is required if _isnan() returns a nonzero value (TRUE); if it returns 0 (FALSE), then it is not possible to know the correct Endian format without further information.

How to read data which is written in little endian using labview

dear all
My program in c++ creats a binary file with data as
struct varaiable
int a;
double b;
doublec;
float d[30];
floate[50];
char time_date[30];
30 such data type is stored in one .b file.
when i am reading the file back using labview the data is not the same as stored.
i think that my c++ program stores data in little endian format while labview is retrieving it in big endian format.
i Checked and found that when i am storing a=2 then while retrieving back in lab i displays it as HEX of 02 00 00 00.
Same is the problem with double and float.so can somebody tell me how i can change labview to read in little endian format.
I am using labview7.1
Abhimaniu

Yes you are correct. Labview and C uses different endian format. Do not worry you have the toolbox in labview. The most important is the typecast function. With this you can convert anything into everything. And as long as you keep the number of bytes and the internal order intact you will always be able to convert things back. I have made an example for you regarding this topic. It is in LV7.1. I have kept this version since the new butt ugly graph cursor introduced in LV8 is a gigantic bug, and should never have seen the light of day. Well enough of my frustration. In the data manipulation palette you will find the type cast and tools for byte/integer swapping and byte/integer splitting and merging. If have done a similar thing before. But I do not remember exactly how to convert a C double into a labview DBL. But take a look at the number 123.123. In this number all the internal bytes are different so by comparing the C version and the labview version byte by byte you will be able to convert it correct. I your case the C struct is 4+8+8+(8X30)+(8x50)+30=690 bytes long. So this is the number of bytes you read each time. I your case I think it will be more convenient to read it as an array of 345 I16(words) but this is up to you.
Besides which, my opinion is that Express VIs Carthage must be destroyed deleted
(Sorry no Labview "brag list" so far)
Attachments:
sample.vi ‏59 KB

TCP Programming / Why do not need to worry about Big and little Endian?

Please help, I do not understand this concept please explain.
The architecture of a CPU is either little-endian or big-endian; some modern CPU's allow a choice via software.
The TCP/IP protocol standard specifies that all the bytes that make up an item must be sent in "network order", which happends to be big-endian. Intel Pentium CPU's are little-endian.
This implies that on an Intel machine the TCP software will have to chop an int into bytes and then reverse the bytes before transmitting them.
Why does the JAVA TCP software does not need to perform the reversal?
Thanks,
Alex

But why would I need to use the DataOutputStream,You don't have to.
But that's what the Java API provides for sending java primitives over a stream. You wouldn't have to use that. You could chop the int into bytes yourself, and send the bytes, and your Java code still wouldn't have to worry about the endiannes of it, because the VMs handle that.
DataOutputStream just does the chopping and reassembling for you, so it's easier than doing it yoursefl.

Why do not need to worry about Big and little Endian in Java

Please help, I do not understand this concept please explain.
The architecture of a CPU is either little-endian or big-endian; some modern CPU's allow a choice via software.
The TCP/IP protocol standard specifies that all the bytes that make up an item must be sent in "network order", which happends to be big-endian. Intel Pentium CPU's are little-endian.
This implies that on an Intel machine the TCP software will have to chop an int into bytes and then reverse the bytes before transmitting them.
Why does the JAVA TCP software does not need to perform the reversal?
Thanks,
Alex

Java doesn't give you direct access to the individual bytes of a larger data item such as an integer. For this reason you don't have the usual endian problems that occur in C. The actual handling of this is in the DataOutputStream and DataInputStream where the integer is coverted to and from bytes using arithmetic, not by fiddling with the internal structure.
Note that regardless of the machine architecture the operation value%256 will return the low order 8 bits. It's less efficient than assigning an int* to a char*, but it's not fraught with the endian problems or any of the other hardware baggage.

How to find machine is Big/little endian?

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