ericliu

@eric.liu.developer

Java bits: 0xFF and 0xFFL

Some Random Binaries

I wrote a method to read a long typed number from an InputStream. The code is as follows:

public static long readLong(final ByteArrayInputStream inputStream) {
long n = 0L;
n |= ((inputStream.read() & 0xFF) << 0);
n |= ((inputStream.read() & 0xFF) << 8);
n |= ((inputStream.read() & 0xFF) << 16);
n |= ((inputStream.read() & 0xFF) << 24);
n |= ((inputStream.read() & 0xFF) << 32);
n |= ((inputStream.read() & 0xFF) << 40);
n |= ((inputStream.read() & 0xFF) << 48);
n |= ((inputStream.read() & 0xFF) << 56);
return n;
}

It returns wrong results and I was scratching my head around what went wrong.

After digging into it, I found that the method java.io.InputStream#read() returns an int type. When a very big left bit shift(like << 32)is performed on an int, the bits will go over bound and the high bits will be discarded.

The solution is to use a long type 0xFFL when we are performing the & operation on an int while hoping to get a long type as the result of such operations:

i & 0xFFL

The programme will perform a sign extension on i and keep all the bits.

So the correct code should look as follows:

public static long readLong(final ByteArrayInputStream inputStream) {
long n = 0L;
n |= ((inputStream.read() & 0xFFL) << 0);
n |= ((inputStream.read() & 0xFFL) << 8);
n |= ((inputStream.read() & 0xFFL) << 16);
n |= ((inputStream.read() & 0xFFL) << 24);
n |= ((inputStream.read() & 0xFFL) << 32);
n |= ((inputStream.read() & 0xFFL) << 40);
n |= ((inputStream.read() & 0xFFL) << 48);
n |= ((inputStream.read() & 0xFFL) << 56);
return n;
}

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