问题
This is what I see in java, and it puzzles me.
Long.toHexString(0xFFFFFFFF) returns ffffffffffffffff
Similarly, 0xFFFFFFFF and Long.parseLong("FFFFFFFF", 16) are unequal.
回答1:
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).
回答2:
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)
回答3:
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].
回答4:
0xFFFFFFFF is an int literal. When using ints (32 bit in Java) 0xFFFFFFFF equals -1. What your code does:
- the compiler parses
0xFFFFFFFFas anintwith value-1 - the java runtime calls
Long.toHexString(-1)(the-1get "casted" automatically to alongwhich 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"
来源：https://stackoverflow.com/questions/9065727/why-long-tohexstring0xffffffff-returns-ffffffffffffffff