C++ 将 UINT16 值转换为 UINT8 数组 [2]
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Converting a UINT16 value into a UINT8 array[2]
提问by dp.
This question is basically the second half to my other Question
这个问题基本上是我另一个问题的后半部分
How can I convert a UINT16 value, into a UINT8 * array without a loop and avoiding endian problems.
如何在没有循环的情况下将 UINT16 值转换为 UINT8 * 数组并避免字节序问题。
Basically I want to do something like this:
基本上我想做这样的事情：
UINT16 value = 0xAAFF;
UINT8 array[2] = value;
The end result of this is to store value into a UINT8 array while avoiding endian conversion.
这样做的最终结果是将值存储到 UINT8 数组中，同时避免字节序转换。
UINT8 * mArray;
memcpy(&mArray[someOffset],&array,2);
When I simply do memcpy with the UINT16 value, it converts to little-endian which ruins the output. I am trying to avoid using endian conversion functions, but think I may just be out of luck.
当我简单地使用 UINT16 值执行 memcpy 时，它会转换为会破坏输出的 little-endian。我试图避免使用字节序转换函数，但我想我可能只是运气不好。
回答by Martin B
How about
怎么样
UINT16 value = 0xAAFF;
UINT8 array[2];
array[0]=value & 0xff;
array[1]=(value >> 8);
This should deliver the same result independent of endianness.
这应该提供与字节顺序无关的相同结果。
Or, if you want to use an array initializer:
或者，如果您想使用数组初始值设定项：
UINT8 array[2]={ value & 0xff, value >> 8 };
(However, this is only possible if valueis a constant.)
（但是，这只有在value是常数时才有可能。）
回答by Christoph
There's no need for conditional compilation. You can bit-shift to get the higher and lower byte of the value:
不需要条件编译。您可以进行位移以获取值的高字节和低字节：
uint16_t value = 0xAAFF;
uint8_t hi_lo[] = { (uint8_t)(value >> 8), (uint8_t)value }; // { 0xAA, 0xFF }
uint8_t lo_hi[] = { (uint8_t)value, (uint8_t)(value >> 8) }; // { 0xFF, 0xAA }
The casts are optional.
演员表是可选的。
回答by Jonathan
Assuming that you want to have the high-order byte above the lower-order byte in the array:
假设您希望数组中的高位字节高于低位字节：
array[0] = value & 0xff;
array[1] = (value >> 8) & 0xff;
回答by Niki Yoshiuchi
union TwoBytes
{
UINT16 u16;
UINT8 u8[2];
};
TwoBytes Converter;
Converter.u16 = 65535;
UINT8 *array = Converter.u8;
回答by omikes
I used this thread to develop a solution that spans arrays of many sizes:
我使用这个线程开发了一个跨越多种大小数组的解决方案：
UINT32 value = 0xAAFF1188;
int size = 4;
UINT8 array[size];
int i;
for (i = 0; i < size; i++)
{
array[i] = (value >> (8 * i)) & 0xff;
}
回答by RightFullRudder
A temporary cast to UINT16* should do it:
临时转换为 UINT16* 应该这样做：
((UINT16*)array)[0] = value;
