关于枚举和按位运算
也许问题是如此简单......
有一个枚举定义：
enum uop_flags_enum {
FICOMP = 0x001,
FLCOMP = 0x002,
FFCOMP = 0x004,
FMEM = 0x008,
FLOAD = 0x010,
FSTORE = 0x020,
FCTRL = 0x040,
FCALL = 0x080,
FRET = 0x100,
FCOND = 0x200
};
代码中的某个地方有：
if (uop->flags & FCTRL)
这个条件何时为真，何时不为真？
Maybe the question is so simple...
There is an enum definition:
enum uop_flags_enum {
FICOMP = 0x001,
FLCOMP = 0x002,
FFCOMP = 0x004,
FMEM = 0x008,
FLOAD = 0x010,
FSTORE = 0x020,
FCTRL = 0x040,
FCALL = 0x080,
FRET = 0x100,
FCOND = 0x200
};
Somewhere in the code there is:
if (uop->flags & FCTRL)
When this condition is true and when it is not?
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最终，此代码检查 uop->flags 变量中的单个位（FCTRL 标志）是否打开。
但这里有一些解释：
代码
if(X)隐式地检查 X 是否为“true”值。对于整数，0 是唯一的“假”值，其他所有值都是“真”。
因此，您的代码相当于：
if (0 != (uop->flags & FCTRL))现在，这是什么意思？
&运算符执行“按位与”，这意味着左侧的每个位与右侧的相应位进行与运算。因此，如果我们以二进制写出两个操作数：
在本例中，如果对每对位执行“AND”，您将得到结果：
其计算结果为 false，实际上在该示例中，uop-> ;flags 值没有设置 FCTRL 标志。
现在这是另一个示例，其中设置了标志 is：
相应的 AND 运算结果：
该结果非零，因此为“true”，触发您的
if语句。Ultimately, this code is checking if a single bit (the FCTRL flag) is turned on in the
uop->flagsvariable.But here's some explanation:
Implicitly, the code
if(X)checks for X being a "true" value.For integers, 0 is the only "false" value and everything else is "true".
Therefore your code is equivalent to:
if (0 != (uop->flags & FCTRL))Now, what does that mean?
The
&operator performs a "bitwise AND", which means each bit of the left-hand-side is ANDed with the corresponding bit on the right-hand-side.So if we wrote out our two operands in binary:
In this example, if you perform an "AND" on each pair of bits, you get the result:
Which evaluates to false, and indeed in that example the
uop->flagsvalue does not have the FCTRL flag set.Now here's another example, where the flag is set:
The corresponding ANDed result:
This result is non-zero, therefore "true", triggering your
ifstatement.这是一个枚举，用于定义操作的多个“标志”。您可以通过以下事实来推断这一点：每个定义的值都是 2 的精确幂，因此由值的单个位（“标志”）表示。
这种类型枚举的优点是，您可以使用按位或组合任意数量的标志：
您给出的使用按位与的条件
当且仅当
FCTRL 标志被设置，即当第 7 个uop->flags位被设置。这是因为 FCTRL == 0x040 == 二进制 01000000。This is an enum used to define a number of "flags" for an operation. You can deduce this by the fact that every defined value is an exact power of two, and because of this is represented by a single bit ("flag") of a value.
The advantage of this type of enum is that you can combine as many of the flags as you want by using bitwise OR:
The condition you give, which uses bitwise AND
is true if and only if when the
FCTRLflag is set, i.e. when the 7th bit ofuop->flagsis set. This is because FCTRL == 0x040 == binary 01000000.当uop->flags中对应于FCTRL(0x040)的位被设置时，条件为真。 '&'是按位与，实际上屏蔽了除 FCTRL 设置的位之外的所有位。
When the bit corresponding to FCTRL (0x040) is set in uop->flags, the condition is true. '&' is a bitwise AND in effect masking all bits but the ones set by FCTRL.
当该位被设置时，条件为真。 0x40 是 1000000，因此当
flags中的第 7 位被设置时 - 它将为 true。The condition is true when the bit is set. 0x40 is 1000000, so when the 7th bit in
flagsis set - it will be true.由于枚举类型使用二进制数字的位置（即单位、2、4、8、16 等），并且操作执行逻辑与。如果设置了该位，则该值将不会为零（真），否则将为假。
As the enumerate type is making use of the position of binary digits (i.e. units, 2, 4, 8, 16 etc) and the operation does a logic and. If that bit place is set the value will not be zero (true) otherwise it will be false.
在这种情况下，每个下一个枚举项都会左移 1 位，因此只需检查
variable & 是否设置了某些标志即可合法地检查是否设置了某个标志。标志==真。但是，如果我们想设置多位标志模式怎么办？例如-何时如何检查我们的变量 X 是否设置了此标志？我们不能只是做
X & myFlag == true，因为例如0b1000 & myFlag == true和0b0010 & myFlag == true- 但 0b1000 和 0b0010 都没有设置我们的两个位！出于这个原因，我更喜欢对位掩码进行全面检查，它可以在枚举中定义多位模式：hth！
In this case each next enum item is shifted by 1 bit left, so it is legal to check if some flag is set by just checking if
variable & flag == true. However what if we want to set multi-bit flag pattern ? For example-when how to check if our variable X has this flag set ? We can't just do
X & myFlag == true, because for example0b1000 & myFlag == trueand0b0010 & myFlag == true- but neither 0b1000 nor 0b0010 has our TWO bits set ! For this reason I prefer full checking of bitmask which lets to define multi-bit patterns in enum:hth!