Bit shifting, again!

Forum 2005-2010 (read only) Software Syntax & Programs
1

Hey guys,

Still learning here, and not great at these bitwise operations. I have a protocol that specifies data be sent as 2 bytes, however the 2 most significant bits of any byte are reserved. That is, the two first bits must always be zero. So 00111111 is valid, but 11111111 is not.

That said, they're still 2-byte values. So how do I spit out a value like this for any given int? Up to 4095, I presume (12 bits total). Say I want to send 1000 for example, that would be 00000111 00101000. How do I do this programmatically in code?

Thanks!

2

Would I do this?

int a = 1000;

int MSB = a&0x0fc0 >> 6;
int LSB = a&0x003f;

Or something like that?

Thanks!

3

I think your way will work. As a matter of style, I would do it this way:

const int lower_6_bits = B00111111;
byte MSB = (a >> 6) & lower_6_bits;
byte LSB = a & lower_6_bits;

If you do the shift first, the mask is the same for both bytes.

-Mike

4

You're right, that's definitely clearer. Thanks!

5

To get MSB (top 8 bits) from a 16-bit variable you need to shift 8 bits right (not 6).

const unsigned int a = 1000;
byte MSB = (a >> 8) & 0x3f;
byte LSB = a & 0x3f;

Also observe type of MSB/LSB - they should be byte (8 bits).

6

BenF said, "To get MSB (top 8 bits) from a 16-bit variable you need to shift 8 bits right (not 6). "

My first reaction was, oops! He's right. Then I realized that we weren't trying to get the top 8 bits from a 16-bit variable, we were trying to get the top two bits of the least-significant byte and the bottom six bits of the most-significant byte. That's because we are trying to output the lower 12 bits of the variable, split into two bytes, with each byte having 6 of the 12 bits.

Each letter represents one bit, "A" is the least significant bit:

PONMLKJI HGFEDCBA

The low order byte we send out will have the bits:

FEDCBA

For the high order byte, we want to send:

LKJIHG

So we need to shift it down 6 positions.

-Mike

P.S. You're absolutely right that the data type for LSB and MSB needs to be "byte".

7

That's because we are trying to output the lower 12 bits of the variable, split into two bytes, with each byte having 6 of the 12 bits.

In that case you're doing just fine! :slight_smile:

As an exercise then perhaps you want to have a go at doing the reverse.

unsigned int a;
byte MSB = B00000111;
byte LSB = B00101000;

// combine lower 6 bits from MSB/LSB into lower 12-bit of a
a = ?

If you get this right - you will qualify as a post graduate bit/shifter :wink:

8

Too easy :slight_smile:

unsigned int a;
byte MSB = B00000111;
byte LSB = B00101000;

// combine lower 6 bits from MSB/LSB into lower 12-bit of a
a = ((unsigned int)MSB << 6) | LSB;

I'm not sure if the type cast is necessary, but it doesn't hurt anything and removes any doubt.

Regards,

-Mike

9

Too easy

This looks good!

I would always include the 16-bit typecast because as you suggests the result (potential bit loss) would otherwise depend on the compiler - and we don't want that (an alternative safe approach is to first move MSB to 'a' and then do the left shift).

If we don't know the value of MSB/LSB I would aslo add code to force the two upper bits to '0' as in the following:

MSB &= B00111111;
LSB &= B00111111;