为什么Long.toHexString(0xFFFFFFFF)返回ffffffffffffffff [英] Why Long.toHexString(0xFFFFFFFF) returns ffffffffffffffff
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问题描述
这是我在Java中看到的，这使我感到困惑.
This is what I see in java, and it puzzles me.
Long.toHexString(0xFFFFFFFF)返回 ffffffffffffffff
类似地， 0xFFFFFFFF 和 Long.parseLong("FFFFFFFF"，16)不相等.
推荐答案
正如其他人所说， 0xFFFFFFFF 的值为 int 值 -1 ，它被提升为 long .
As others have said, 0xFFFFFFFF evaluates to the int value -1, which is promoted to a long.
要获得预期的结果，请用 L 后缀限定常量，以指示应将其视为 long ，即 Long.toHexString(0xFFFFFFFFL).
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).
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