C++ 如何使用高字节和低字节？
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How to use high and low bytes?
提问by Sterling
I am trying to represent 32768 using 2 bytes. For the high byte, do I use the same values as the low byte and it will interpret them differently or do I put the actual values? So would I put something like 32678 0 or 256 0? Or neither of those? Any help is appreciated.
我正在尝试使用 2 个字节来表示 32768。对于高字节，我是否使用与低字节相同的值，它会以不同的方式解释它们还是我放置实际值？那么我会输入 32678 0 或 256 0 之类的东西吗？或者两者都没有？任何帮助表示赞赏。
回答by John McFarlane
In hexadecimal, your number is 0x8000 which is 0x80 and 0x00.
To get the low byte from the input, use low=input & 0xffand to get the high byte, use high=(input>>8) & 0xff.
在十六进制中，您的号码是 0x8000，即 0x80 和 0x00。要从输入中获取低字节，请使用low=input & 0xff并获取高字节，请使用high=(input>>8) & 0xff.
Get the input back from the low and high byes like so: input=low | (high<<8).
获得从低输入背部和高轮空，像这样：input=low | (high<<8)。
Make sure the integer types you use are big enough to store these numbers. On 16-bit systems, unsigned int/shortor signed/unsigned longshould be be large enough.
确保您使用的整数类型足够大以存储这些数字。在 16 位系统上，unsigned int/short或signed/unsigned long应该足够大。
回答by Ignacio Vazquez-Abrams
Bytes can only contain values from 0 to 255, inclusive. 32768 is 0x8000, so the high byte is 128 and the low byte is 0.
字节只能包含 0 到 255（含）之间的值。32768 是 0x8000，所以高字节是 128，低字节是 0。
回答by cyberponk
Pointers can do this easily, are MUCH FASTER than shifts and requires no processor math.
指针可以很容易地做到这一点，比移位快得多，并且不需要处理器数学。
BUT: If I understood your problem, you need up to 32768 stored in 2 bytes, so you need 2 unsigned int's, or 1 unsigned long. Just change int for long and char for int, and you're good to go.
但是：如果我理解你的问题，你最多需要 32768 个存储在 2 个字节中，所以你需要 2 个 unsigned int 或 1 个 unsigned long。只需将 int 更改为 long 并将 char 更改为 int，您就可以开始了。
回答by Ernesto A. Garrote
Try this function. Pass your Hi_Byte and Lo_Byte to the function, it returns the value as Word.
试试这个功能。将您的 Hi_Byte 和 Lo_Byte 传递给该函数，它会将值作为 Word 返回。
WORD MAKE_WORD( const BYTE Byte_hi, const BYTE Byte_lo)
{
return (( Byte_hi << 8 ) | Byte_lo & 0x00FF );
}
回答by Alnitak
32768 is 0x8000, so you would put 0x80 (128) in your high byte and 0 in your low byte.
32768 是 0x8000，因此您可以将 0x80 (128) 放在高字节中，将 0 放在低字节中。
That's assuming unsigned values, of course. 32768 isn't actually a legal value for a signed 16-bit value.
当然，这是假设无符号值。32768 实际上不是有符号 16 位值的合法值。
回答by azyoot
32768 in hex is 0080 on a little-endian platform. The "high" (second in our case) byte contains 128, and the "low" one 0.
在小端平台上，十六进制的 32768 是 0080。“高”（在我们的例子中是第二个）字节包含 128，而“低”字节包含 0。
