Re: problems reading binary file
According to Mikee <mikee.read@googlemail.com>:
DataInputStream in = new DataInputStream(new
BufferedInputStream(new FileInputStream(filename)));
System.out.print(ByteSwapper.swap(in.readFloat())+" ");
System.out.print(ByteSwapper.swap(in.readFloat())+" ");
System.out.print(ByteSwapper.swap(in.readFloat())+" ");
System.out.print(ByteSwapper.swap(in.readFloat())+" ");
System.out.println(ByteSwapper.swap(in.readFloat())+" ");
[...]
public static float swap (float value)
{
int intValue = Float.floatToIntBits (value);
intValue = swap (intValue);
return Float.intBitsToFloat (intValue);
}
This is conceptually flawed. What you do, here, is the following:
1. read four bytes
2. interpret them as a float (in.readFloat())
3. convert them back to four bytes, as an "int" value
(Float.floatToIntBits())
4. byte-swap the bytes
5. finally reinterpret the bytes as a float (Float.intBitsToFloat())
Steps 2 and 3 basically cancel each other: step 3 undoes what step 2
did. The trouble is that this roundtrip (steps 2 and 3) may lose some
information, because some bit patterns encode a "NaN" (a special value
for floats) and Float.floatToIntBits() normalizes NaN values. This is
probably what happens to you: when read "byte-swapped", your -99.99999
becomes a pattern for a NaN, which gets normalized to another NaN,
thus changing the bit pattern.
You _might_ correct your code by using Float.float.toRawIntBits()
instead of Float.floatToIntBits(), but even if it works, it is
non-robust (it assumes that DataInputStream also does things that way).
A more correct way of doing things is to read the bytes as an "int"
directly, which basically removes steps 2 and 3 altogether. This would
look like this:
System.out.print(Float.intBitsToFloat(swap(in.readInt())) + " ");
--Thomas Pornin