byte[] array to uint16_t

Projects Programming
1

Hello,

I am working with an Arduino Leonardo ETH. I am receiving data via ethernet:

...
if ( (client) && (client.available()) > 5)
{
	byte buf[4] = { 0,0,0,0 };

	client.readBytesUntil
	client.readBytes(buf, 4); 
	proccessArray(buf);

	Ethernet.maintain();
}
...

So far, so good. However, now the last two bytes of my buffer array should be converted to a uint16_t. The data is sent with the most significant byte first (big-endian order), but in principle I could change that as I am writing the sending app too. However, if possible, I'd keep it like that.

Now how do I convert that to an uint16_t?

I know, it's quite basic stuff, but I am chemist doing automatization... So hopefully it's forgivable... :slight_smile:

Thanks!

Best,
Andreas

2

uint16_t myInt = buf[3] << 8 | buf[4];  // shift one byte by 8 and OR the two to get your result .

kind-of-thing...

but I'm not sure if the endianness is correct...

3

Hello,

thanks for the help: To test if the endianness is correct, I can just reverse buf[3] and buf[4] in the code, right?

And: It should be byte[3] and byte[2], right? Indices start with zero...?

Thanks,
Andreas

4

andreas_:
Hello,
And: It should be byte[3] and byte[2], right? Indices start with zero...?

Thanks,
Andreas

yeah... I'm in and out of Lua these days :frowning:

5

Well...I better ask. I have to do quite a lot of things in MATLAB and there indices do start with 1...

I hope it's OK to ask a follow up question: Reversing the same thing, meaning creating a byte[2] of an uint16_t would then be:

 uint16_t val = 65535;

  byte low = val & 0xff;
  byte high = (val >> 8) & 0xff;

  byte[2] = {high, low}; // Order for Big Endian;

Or is there a more efficient way?

Best,
Andreas

6

andreas_:

  byte high = (val >> 8) & 0xff;

Or is there a more efficient way?

the low order bits of val end up pushed off into the trash bin, so no need to & 0xFF...