什么是 0xFF？为什么它移位 24 次？
#define SwapByte4(ldata) \
(((ldata & 0x000000FF) << 24) | \
((ldata & 0x0000FF00) << 8) | \
((ldata & 0x00FF0000) >> 8) | \
((ldata & 0xFF000000) >> 24))
0x000000FF 代表什么？我知道十进制的15用十六进制表示为F，但是为什么它<< 24？
#define SwapByte4(ldata) \
(((ldata & 0x000000FF) << 24) | \
((ldata & 0x0000FF00) << 8) | \
((ldata & 0x00FF0000) >> 8) | \
((ldata & 0xFF000000) >> 24))
What does that 0x000000FF represent? I know that decimal 15 is represented in hex as F, but why is it << 24?
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这是一个十六进制值 0x12345678，写为二进制，并用一些位位置进行注释：
...这里是 0x000000FF：
因此按位 AND 仅选择原始值的底部 8 位：
...并将其左移24 位将其从底部 8 位移动到顶部：
...十六进制为 0x78000000。
其他部分处理输入的剩余 8 位部分：
0x12345678 & 0x000000FF ---------- 0x00000078 << 24 = 0x78000000 (as shown above) 0x12345678 & 0x0000FF00 ---------- 0x00005600 << 8 = 0x00560000 0x12345678 & 0x00FF0000 ---------- 0x00340000 >> 8 = 0x00003400 0x12345678 & 0x00000000 ---------- 0x12000000 >> 24 = 0x00000012 | ---------- 0x78563412因此总体效果是将 32 位值 ldata 视为四个 8 位字节的序列，并反转它们的顺序。
Here is a hex value, 0x12345678, written as binary, and annotated with some bit positions:
...and here is 0x000000FF:
So a bitwise AND selects just the bottom 8 bits of the original value:
...and shifting it left by 24 bits moves it from the bottom 8 bits to the top:
...which is 0x78000000 in hex.
The other parts work on the remaining 8-bit portions of the input:
0x12345678 & 0x000000FF ---------- 0x00000078 << 24 = 0x78000000 (as shown above) 0x12345678 & 0x0000FF00 ---------- 0x00005600 << 8 = 0x00560000 0x12345678 & 0x00FF0000 ---------- 0x00340000 >> 8 = 0x00003400 0x12345678 & 0x00000000 ---------- 0x12000000 >> 24 = 0x00000012 | ---------- 0x78563412so the overall effect is to consider the 32-bit value
ldataas a sequence of four 8-bit bytes, and reverse their order.这种代码通常用于在大端和小端格式之间交换内容。还有一个小技巧，可以将某种已知格式（例如小尾数法）的单词转换为当前机器恰好采用的任何尾数法，反之亦然。这会是这样的：
这有效（假设我没有搞砸），因为无论字节实际如何存储，左移都保证向更有效的位移动。转换为 char* 允许您按照字节实际存储在内存中的顺序访问字节。使用此技巧，您不需要检测机器字节顺序即可以已知格式读/写内容。诚然，您也可以只使用标准函数（hton 等）：P
（注意：您必须小心一点，在移动之前转换字符，否则它会溢出您的鞋子。此外， += 不是唯一的选项， |= 可能更有意义，但如果您不习惯的话可能不太清楚，我不确定）
This kind of code tends to be used to swap things between big endian and little endian format. There is also a little trick that will convert a word in some known format (lets say, little endian) into whatever endianness the current machine happens to be, and vice versa. That would go something like this:
This works (assuming I haven't messed it up) because regardless of how bytes are actually stored, shift left is guaranteed to shift towards more significant bits. Converting to a char* allows you to access the bytes in the order they are actually stored in memory. Using this trick you don't need to detect the machine endianness to read/write stuff in a known format. Admittedly you could also just use the standard functions (hton etc.) :P
(Note: You have to be a little careful and cast the char before shifting, otherwise it just overflows all over your shoes. Also, += isn't the only option, |= would probably make more sense but might be less clear if you aren't used to it, I'm not sure)
您需要将
0x000000FF视为位掩码，即在其为1的位置将采用ldata的值，在其为0 - 0的位置将采用该值。为了理解位掩码，您需要将其转换为二进制，使用十六进制很容易，每个十六进制数都是 4 个二进制数字，即：
十六进制 0 = 二进制 0000
十六进制 1 = 二进制 0001
等等。
现在进行移位：请注意，移位从源中获取一些数据（精确的 8 位），并将其移动到目标中的另一个位置。
现在请注意，有
|即所有位掩码 AND 运算上的 OR 运算，即零将保持为零，如果存在“1”，结果将包含 1。希望它有帮助:)
You need to look at the
0x000000FFas a bitmask, i.e. where it's 1 the value ofldatawill be taken and where it's 0 - 0 will be taken.In order to understand the bitmask u need to convert it to binary, with hex it's very easy, every hex number is 4 binary digits, i.e.:
hex 0 = binary 0000
hex 1 = binary 0001
and so on.
Now to shifts: notice that the shift takes some data from the source, 8 bits exactly, and moves it to another location in the destination.
Now note that there's
|i.e. OR operation on all the bitmask AND operations, i.e. zeroes will stay zeroes and in case there's '1' the result will contain one.Hope it helps :)
假设数据是一个 32 位数字，表示为 0x12345678（每个数字为 4 位十六
进制 ） 0x000000FF 表示仅保留最后 8 位（称为位掩码）
= 0x00000078
<< 24 表示将该值向左移动 24 位（78 从位置 24 [0 索引] 开始）
= 0x78000000
|表示逻辑或在本例中只是加法
最终结果 = 0x78563412
读取逻辑操作
Let's say data is a 32 bit number represented as 0x12345678 (each number is 4 bits in hex)
Data & 0x000000FF means keep only the last 8 bits (called a bit mask)
= 0x00000078
The << 24 means move this value to the left 24 bits (78 starts at position 24 [0 index])
= 0x78000000
The | means logical or which in this case will just be an addition
Final result = 0x78563412
Read on logical manipulations