Ashish wrote:
>
> Answer is simple
>
> Java program expects file bytes in big-endian format, while C uses little
> endian
Sorry, wrong. C uses the native byte order of the system. I.e. it
depends on the processor.
x86 (Intel, AMD) => little endian
PowerPC, Alpha, Sparc => big endian
Matthias
> 2175 = > 00001000 01111111
>
> so java will interpret after assuming it's big endian will be 01111111
> 00001000 => 32520
>
> -Ashish
>
> ----- Original Message -----
> From: "Joaquin Rapela" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Wednesday, March 21, 2001 9:56 AM
> Subject: readInt() & byte order
>
> > Hello,
> >
> > I have a binary file containing short integers. I wrote a C program to
> read
> > the file and it works as expected. I wrote a java program to read the file
> > using a DataInputStream and its readInt() method and it is reading the
> short
> > integers in inverted order Instead of a 2175 (100001111111) I get a 32520
> > (111111100001000). To read the integers I read the following code:
> >
> > File inputFile = new File(filename);
> > in = new DataInputStream(new FileInputStream(filename));
> > short current = -1;
> > current = in.readShort();
> >
> > Is there something wrong with this? Shall I use another method?
> >
> > Thanks in advance, Joaquin
>
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--
Matthias Pfisterer <mailto:[EMAIL PROTECTED]>
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