What is best way to convert two uint8 bytes to uint16 byte in arduino?

Question

I'm using Arduino with ESP32 and I'm still at the beginner level. I have two bytes which are uint8 and are variables, I want to convert them to uint16. Which way would be best to do it? I think bitshift should be good

uint8_t first = 0xFF; //lets assume its variable, not constant
uint8_t second = 0xEE;
uint16_t combined = (first << 8) + second;
Serial.println(combined, HEX); // prints FFEE

The code above does the job, but I'm not sure is it the right/best way.

I would use either an arithmetic expression: `(first * 256) + second`, or a bit manipulation expression: `(first << 8) | second`. — Weather Vane, Dec 22 '19 at 19:15
I would change the `+` into a `|`, but that's just for clarity of meaning (i.e., dealing with bits). — Arkku, Dec 22 '19 at 19:21
If on a 16-bit Arduino, use `(first * 256u) + second` to avoid UB of `(first << 8) + second`. — chux - Reinstate Monica, Dec 22 '19 at 19:29
@Ry- With 16-bit `int` and `first << 8`, `first` is promoted to `int`, then shifted. With `first >= 128`, shifting into the sign bit is UB per 6.5.7 4. — chux - Reinstate Monica, Dec 22 '19 at 22:24
@WeatherVane "shifting unsigned values is well defined?" unclear. Note: `first << 8` is shifting an `int` due to _integer promotions_. — chux - Reinstate Monica, Dec 22 '19 at 22:28
@chux-ReinstateMonica: Huh, for some reason I thought a newer C standard made that implementation-defined. Thanks :) — Ry-, Dec 22 '19 at 22:33
@Ry- IAC, on review, OP's ESP32 certainly uses a 32-bit `int`, in which case, [my concern](https://stackoverflow.com/questions/59447181/what-is-best-way-to-convert-two-uint8-bytes-to-uint16-byte-in-arduino?noredirect=1#comment105077414_59447181) does not apply here. — chux - Reinstate Monica, Dec 22 '19 at 22:35

score 1 · Accepted Answer · answered Dec 22 '19 at 23:06

Which way would be best to do it? I think bitshift should be good

With a 32-bit int/unsigned, all below are the same function and a compiler can emit the same code.

uint8_t first, second;
...
uint16_t combined = (first << 8) + second;
uint16_t combined = (first << 8) | second;
uint16_t combined = (first << 8) ^ second;
uint16_t combined = (first * 256) + second;
uint16_t combined = (first * 256) | second;
....

What is "best" is likely "any of the above" given the compiler is a good one.

For me, I would code as the below if I felt the compiler is potentially weak.

uint16_t combined = ((unsigned)first << 8) | second;

Else I would code to what makes sense for the larger code and clarity. If arithmetic in nature, then the following:

uint16_t combined = (first * 256u) + second;  // recommended

Lastly I might try a union trick to compel a weak compile, but such micro-optimizations are of dubious performance benefits vs. maintenance effort.

With a 16-bit int/unsigned, best to avoid int overflow.

uint16_t combined = ((uint16_t)first << 8) | second;
uint16_t combined = (first * 256u) + second;

Thanks a lot, thats what I wanted to know – TheAfrizzz Dec 23 '19 at 03:28 — TheAfrizzz, Dec 23 '19 at 03:28

What is best way to convert two uint8 bytes to uint16 byte in arduino?

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