Why Long.toHexString(0xFFFFFFFF) returns ffffffffffffffff

Question

This is what I see in java, and it puzzles me.

Long.toHexString(0xFFFFFFFF) returns ffffffffffffffff

Similarly, 0xFFFFFFFF

Answer 1

This:

Long.toHexString(0xFFFFFFFF)

is equivalent to:

Long.toHexString(-1)

which is equivalent to:

Long.toHexString(0xFFFFFFFFFFFFFFFFL)

Basically, the problem is that you're specifying a negative int value, which is then being converted to the equivalent negative long value, which consists of "all Fs". If you really want 8 Fs, you should use:

Long.toHexString(0xFFFFFFFFL)

Answer 2

0xFFFFFFFF is an int literal. When using ints (32 bit in Java) 0xFFFFFFFF equals -1. What your code does:

• the compiler parses 0xFFFFFFFF as an int with value -1
• the java runtime calls Long.toHexString(-1) (the -1 get "casted" automatically to a long which is expected here)

And when using longs (64 bit in Java) -1 is 0xffffffffffffffff.

long literals are post-fixed by an L. So your expected behaviour is written in Java as:

Long.toHexString(0xFFFFFFFFL)

and Long.toHexString(0xFFFFFFFFL) is "ffffffff"

Answer 3

As others have said, 0xFFFFFFFF evaluates to the int value -1, which is promoted to a long.

To get the result you were expecting, qualify the constant with the L suffix to indicate it should be treated as a long, i.e. Long.toHexString(0xFFFFFFFFL).

Answer 4

Of course, Long in java is 64-bits long! 0xFFFFFFFF means -1 as an int, when written in 64 bits, it's ffffffffffffffff.

However, if the number were unsigned, the string would also be ffffffff [but there's no unsigned in java].