"u00F6"
is not a byte array. It's a string containing a single char. Execute the following test instead:
public static void main(String[] args) throws Exception {
byte[] b = new byte[] {(byte) 0x00, (byte) 0xf6};
String s = new String(b, "ISO-8859-1"); // decoding
byte[] b2 = s.getBytes("ISO-8859-1"); // encoding
System.out.println("Are the bytes equal : " + Arrays.equals(b, b2)); // true
}
To check that this is true for any byte, just improve the code an loop through all the bytes:
public static void main(String[] args) throws Exception {
byte[] b = new byte[256];
for (int i = 0; i < b.length; i++) {
b[i] = (byte) i;
}
String s = new String(b, "ISO-8859-1");
byte[] b2 = s.getBytes("ISO-8859-1");
System.out.println("Are the bytes equal : " + Arrays.equals(b, b2));
}
ISO-8859-1 is a standard encoding. So the language used (Java, C# or whatever) doesn't matter.
Here's a Wikipedia reference that claims that every byte is covered:
In 1992, the IANA registered the character map ISO_8859-1:1987, more commonly known by its preferred MIME name of ISO-8859-1 (note the extra hyphen over ISO 8859-1), a superset of ISO 8859-1, for use on the Internet. This map assigns the C0 and C1 control characters to the unassigned code values thus provides for 256 characters via every possible 8-bit value.
(emphasis mine)
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