按位移位说明
假设我将变量 x 初始化为 425。用二进制表示，即 110101001。
将其向右移位 2，如下所示： int a = x >>> 2;，答案是：106。二进制为 1101010。这是有道理的，因为最右边的两个位被删除，并且两个零被添加到左侧。
将其向左移位 2，如下所示：int a = x << 2;，答案是：1700。在二进制中，这是 11010100100。我不明白这是如何运作的。为什么最左边的两个位被保留？我怎样才能丢掉它们？
谢谢你，
Assume I have the variable x initialized to 425. In binary, that is 110101001.
Bitshifting it to the right by 2 as follows: int a = x >> 2;, the answer is: 106. In binary that is 1101010. This makes sense as the two right-most bits are dropped and two zero's are added to the left side.
Bitshifting it to the left by 2 as follows: int a = x << 2;, the answer is: 1700. In binary this is 11010100100. I don't understand how this works. Why are the two left most bits preserved? How can I drop them?
Thank you,
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这是因为
int在您的系统上可能是 32 位。 （假设 x 是int类型。）因此，您的
425实际上是：当左移 2 时，您会得到：
直到您完全过去，没有任何内容会被移动32.（严格来说，有符号整数的溢出是 C/C++ 中未定义的行为。）
要删除移出的位，您需要对作为数字原始长度的掩码进行按位 AND 操作：
This is because
intis probably 32-bits on your system. (Assuming x is typeint.)So your
425, is actually:When left-shifted by 2, you get:
Nothing gets shifted off until you go all the way past 32. (Strictly speaking, overflow of signed-integer is undefined behavior in C/C++.)
To drop the bits that are shifted off, you need to bitwise AND against a mask that's the original length of your number:
首先，不要移动有符号整数。位运算仅对于无符号整数类型来说是普遍明确的。
其次，如果可以使用
* 4和/ 4为什么要转变？第三，当超出类型的大小时，仅删除左侧的位。如果您想在数学上“在左侧截断”，请执行模运算：
按位相当于与位模式的 AND：
(x << 2) & 0xFF（也就是说，C 中的基本无符号整数类型始终隐式“模 2n”，其中 n 是该类型的位数。）
First off, don't shift signed integers. The bitwise operations are only universally unambiguous for unsigned integral types.
Second, why shift if you can use
* 4and/ 4?Third, you only drop bits on the left when you exceed the size of the type. If you want to "truncate on the left" mathematically, perform a modulo operation:
The bitwise equivalent is AND with a bit pattern:
(x << 2) & 0xFF(That is, the fundamental unsigned integral types in C are always implicitly "modulo 2n", where n is the number of bits of the type.)
为什么你会期望它们被丢弃？你的 int （可能）消耗 4 个字节。你正在将它们转移到它正确占据的空间中。
评估期间会占用内存中的整个 4 字节空间。您需要完全移出内存中的该空间才能“删除”它们。
Why would you expect them to be dropped? Your int (probably) consumes 4 bytes. You're shifting them into a space that it rightfully occupies.
The entire 4-byte space in memory is embraced during evaluation. You'd need to shift entirely out of that space in memory to "drop" them.