I've a few lines of code within a project, that I can't see the value of...
buffer[i] = (currentByte & 0x7F) | (currentByte & 0x80);
It reads the filebuffer from a file, stored as bytes, and then transfers then to buffer[i] as shown, but I can't understand what the overall purpose is, any ideas?
Thanks
As the other answers already stated, (currentByte & 0x7F) | (currentByte & 0x80) is equivalent to (currentByte & 0xFF). The JLS3 15.22.1 says this is promoted to an int:
When both operands of an operator &, ^, or | are of a type that is convertible (§5.1.8) to a primitive integral type, binary numeric promotion is first performed on the operands (§5.6.2). The type of the bitwise operator expression is the promoted type of the operands.
because JLS3 5.6.2 says that when currentByte has type byte and 0x7F is an int (and this is the case), then both operands are promoted to int.
Therefore, buffer will be an array of element type int or wider.
Now, by performing & 0xFF on an int, we effectively map the original byte range -128..127 into the unsigned range 0..255, an operation often used by java.io streams for example.
You can see this in action in the following code snippet. Note that to understand what is happening here, you have to know that Java stores integral types, except char, as 2's complement values.
byte b = -123;
int r = b;
System.out.println(r + "= " + Integer.toBinaryString(r));
int r2 = b & 0xFF;
System.out.println(r2 + "= " + Integer.toBinaryString(r2));
Finally, for a real-world example, check out the Javadoc and implementation of the read method of java.io.ByteArrayInputStream:
/**
* Reads the next byte of data from this input stream. The value
* byte is returned as an <code>int</code> in the range
* <code>0</code> to <code>255</code>. If no byte is available
* because the end of the stream has been reached, the value
* <code>-1</code> is returned.
*/
public synchronized int read() {
return (pos < count) ? (buf[pos++] & 0xff) : -1;
}
(currentByte & 0x7F) | (currentByte & 0x80)
is equivalent to
currentByte & (0x7F | 0x80)
which equals
currentByte & 0xFF
which is exactly the same as
currentByte
Edit: I only looked at the right side of the assignment, and I still think the equivalance is true.
However, it seems like the code wants to cast the signed byte to a larger type while interpreting the byte as unsigned.
Is there an easier way to cast signed-byte to unsigned in java?
I think someone did too much thinking here. That's just not right.
I have but one remark
It's code left behind. Unless you know you're on a fishy run-time? What's the type of the 'buffer' anyway?
The complicated bitwise logic is completely superfluous.
for (int i = 0; i < buffer.length; i++) {
buffer[i] = filebuffer[currentPosition + i] & 0xff;
}
does the same thing. If buffer is declared as an array of bytes you may even leave out the & 0xff, but unfortunately the declaration is not shown.
The reason may be that the original developer was confused by bytes being signed in Java.
The result of a bitwise AND operation has a 1 on that bits where both bits are 1 while the result of a bitwise OR operation hase a on that bits where either one of bot bits is 1.
So an example evaluation for the value 0x65:
01100101 0x65
& 01111111 0x7F
===============
01100101 0x65
01100101 0x65
& 10000000 0x80
===============
00000000 0x00
01100101 0x65
| 00000000 0x00
===============
01100101 0x65
The good thing about these kinds of logical operations: you can try every possible combination (all 256 of them) and verify that you get the answer you expected.
Turns out, the file which the byte was being read from was in a signed bit notation, and of a different length, therefore it was requried to perform this task to allow it to be extended to the java int type, while retaining its correct sign :)
