C语言 将两个 uint8_t 组合为 uint16_t
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Combining two uint8_t as uint16_t
提问by johan
I have the following data
我有以下数据
uint8_t d1=0x01;
uint8_t d2=0x02;
I want to combine them as uint16_tas
我想将它们组合uint16_t为
uint16_t wd = 0x0201;
How can I do it?
我该怎么做？
回答by md5
You can use bitwise operators:
您可以使用按位运算符：
uint16_t wd = ((uint16_t)d2 << 8) | d1;
Because:
因为：
(0x0002 << 8) | 0x01 = 0x0200 | 0x0001 = 0x0201
回答by R.. GitHub STOP HELPING ICE
The simplest way is:
最简单的方法是：
256U*d2+d1
回答by Lundin
This is quite simple. You need no casts, you need no temporary variables, you need no black magic.
这很简单。你不需要强制转换，不需要临时变量，不需要黑魔法。
uint8_t d1=0x01;
uint8_t d2=0x02;
uint16_t wd = (d2 << 8) | d1;
This is always well-defined behavior since d2 is always a positive value and never overflows, as long as d2 <= INT8_MAX.
这始终是明确定义的行为，因为 d2 始终是正值并且永远不会溢出，只要d2 <= INT8_MAX.
(INT8_MAX is found in stdint.h).
（INT8_MAX 可在 stdint.h 中找到）。
回答by fizzbuzz
(uint16_t)((d2 << 8) + (d1 & 0x00ff))
