从4 2位创建一个字节（8位）

Question

I am trying to figure out a way to get as much out of the limited memory in my microcontroller (32kb) and am seeking suggestions or pointers to an algorithm that performs what I am attempting to do.

Some background: I am sending Manchester Encoded bits out a SPI (Serial Peripheral Interface) directly from DMA. As the smallest possible unit I can store data into DMA is a byte (8 bits), I am having to represent my 1's as 0b11110000 and my 0's as 0b00001111. This basically means that for every bit of information, I need to use a byte (8 bits) of memory. Which is very inefficient.

If I could reduce this, so that my 1's are represented as 0b10 and my 0's as 0b01, I'd only have to use a 1/4 of a byte (2 bits) for every 1 bit of memory, which is fine for my solution.

Now, if I could save to DMA in bits, this would not be a problem, but of course I need to work with bytes. So I know the solution to my problem involves collecting the 8 bits (or in my case, 4 2bits) and then storing to DMA as a byte.

Questions:

Is there a standard way to solve this problem?

How can I some how create a 8 bit number from a collection of 4 2 bit numbers? But I do not want the addition of these numbers, but the actual way it looks when collected together.

For example: I have the following 4 2 bit numbers (keeping in mind that 0b10 represents 1 and 0b01 represents 0) (Also, the type these are stored in is open to the solution, as obviously there is no such thing as a 2 bit type)

Number1: 0b01 Number 2: 0b10 Number 3: 0b10 Number4: 0b01

And I want to create the following 8 bit number from these:

8 Bit Number: 0b01 10 10 01 or without the spaces 0b01101001 (0x69)

I am programming in c

Answer 1

It seems that you can pack four numbers a , b , c , d , all of which of value zero or one, like so:

64 * (a + 1) + 16 * (b + 1) + 4 * (c + 1) + (d + 1)

This is using the fact that x + 1 encodes your two-bit integer: 1 becomes 0b10, and 0 becomes 0b01.

Answer 2

It's Manchester encoding so 0b11110000 and 0b00001111 should be the only candidates. If so, then reduce the memory by a factor of 8.

uint8_t PackedByte = 0;
for (i=0; i<8; i++) {
  PackedByte <<= 1;
  if (buf[i] == 0xF0) //  0b11110000
    PackedByte++;
}

Other other hand, if it's Manchester encoding and one may not have perfect encoding, then there are 3 results: 0, 1, indeterminate.

uint8_t PackedByte = 0;
for (i=0; i<8; i++) {
  int upper = BitCount(buf[i] >> 4);
  int lower = BitCount(buf[i] & 0xF);
  if (upper > lower)
    PackedByte++;
  else if (upper == lower)
    Hande_Indeterminate();
}

Various simplifications absent in the above, but shown for logic flow.

Answer 3

To number get abcd from (a,b,c,d) you need to shift the number to their places and OR :-

(a<<6)|(b<<4)|(c<<2)|d