位左移
假设我想将 i 向左移位两次并将值存储在 f 中。
f = i << 2;
这是正确的吗？我到底如何在 C/C++ 中做到这一点？
Let's say I want to bit shift i twice to the left and store the value in f.
f = i << 2;
Is that correct? How exactly do I do this in C/C++?
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是的。
移位在许多位旋转操作中很有用。
这曾经是一个将数字乘以四的好方法。然而，如今，优化编译器往往会为您解决这个问题。
请记住，最左边的两位将被丢弃。
Yes.
Shifts are useful in a number of bit twiddling operations.
This used to be a great way to multiply a number by four. However, these days, optimizing compilers tend to take care of that for you.
Keep in mind that the two leftmost bits are discarded.
作为补充说明：即使您的问题被标记为
C++，但可能值得注意的是，C 和 C++ 在移动负值方面采取了略有不同的路径。在 C++ 中，对负值执行<<或>>的结果是实现定义的。在 C 中，>>是实现定义的，而<<会产生未定义的行为。As an additional note: Even though your question is tagged
C++, it is probably worth noting that C and C++ took slightly different paths with regard to shifting negative values. In C++ the result of doing<<or>>on a negative value is implementation-defined. In C>>is implementation-defined, while<<produces undefined behavior.是的，
i << 2，f = i << 2或f <<= 2都是人们可能想要进行的移位操作。更多转变的事情要记住：
你还有
>>。在位级别，>>对于有符号和无符号类型的工作方式有所不同。<<和>>的优先级低于+和-，这愚弄了一些人，因为人们可能认为它们更像*和/。Yes,
i << 2,f = i << 2, orf <<= 2are all things one might want to do to shift bits.More shift things to keep in mind:
you have
>>as well. At the bit level,>>works differently for signed and unsigned types.the priority of
<<and>>is below that of+and-, which fools some people, as one might imagine them to be more like*and/.为了完整地帮助您进行位运算，您可以查看此页面：uow 教科书 -> bitops.html
For the sake of completeness to help you with your bit operations you can check out this page: uow TEXTBOOK -> bitops.html