java 按位“&”就长了？
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Bitwise "& " on a long?
提问by Regis St-Gelais
I want to obtain the least 32 bits from a number of type long, so I perform bitwise operation "&" on the number with bits mask 0xFFFFFFFF, but the result is not correct, it still contains the other bits.
我想从一个long类型的数字中获取最少的32位，所以我对位掩码为0xFFFFFFFF的数字执行按位运算“&”，但结果不正确，它仍然包含其他位。
for example:
例如：
long a = 0x1234567890ab;
long b = (a & 0xffffffff);
I expect the value of b to be 0x567890ab
我希望 b 的值是 0x567890ab
but in practice, it is still 0x1234567890ab
但实际上，它仍然是 0x1234567890ab
回答by axtavt
Try this:
试试这个：
long a = 0x1234567890ab;
long b = (a & 0xffffffffL);
0xffffffffis a literal of type int, for performing &with longit's promoted to type longby sign extension, therefore it turns into 0xffffffffffffffff. To avoid sign extension you need to write it as a literal of type long: 0xffffffffL.
0xffffffff是 type 的文字int，为了执行&，long它long通过符号扩展被提升为 type ，因此它变成0xffffffffffffffff. 为避免符号扩展，您需要将其写为类型为long:的文字0xffffffffL。
回答by sargas
Does using 0xffffffffL make any difference?
使用 0xffffffffL 有什么不同吗？
I think what happens is that 0xffffffff gets upcasted to a long, and since both int and long are signed it tries to keep the same sign.
我认为发生的情况是 0xffffffff 被向上转换为 long，并且由于 int 和 long 都已签名，因此它会尝试保持相同的符号。
So, since 0xffffffff is -1 as an int, it gets converted to -1 as a long, or 0xffffffffffffffff
因此，由于 0xffffffff 作为 int 是 -1，因此它会被转换为 -1 作为 long，或 0xffffffffffffffff
回答by Scott M.
try doing a bitwise AND with a long value with some leading zeroes instead of a 32 bit value. right now you're basically doing nothing to the number.
尝试使用带有一些前导零而不是 32 位值的长值进行按位 AND 运算。现在你基本上没有对数字做任何事情。
